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Bottled water

The term market research is the “systematic problem analysis, model building and fact finding for the purposes of improved decision making in the marketing of goods and services”, this basically means that through research companies are going to try and give customers what they want. In my assignment I too will use market research, and will be carrying out a research on bottled water. I intend to find out how pupils of my school feel about the idea of bottled water, and how the feel about certain brands, flavours etc.

I also would hope to explain why bottled water has become the success it has, and why pupils of my school prefer it to a bottle filled with ordinary tap water. I have decided to use a questionnaire as part of my primary research. The questionnaire will set out to investigate our schools knowledge of water and we’ll see how many people are consuming it daily. Through the questionnaires I will discover various different aspects in order to find out if it affects the market positively or negatively.

Primary Research is a type of data which you find yourself albeit through…

* Discussions with people

* Direct observation

* Questionnaires and surveys

* Sampling

* Testing through pilot and field trials

For my research I have decided to use the questionnaires and surveys method, combined with a method of sampling. This is to ensure the complete reliability of my research. I have chosen these methods mainly due to the fact that the project is based in school, therefore I must select the quickest methods available, this however will not deflect from the amount of effort needed in order to compile the questionnaire and to distribute them to the correct people.

The two main types of data are collected by market research are…

* Quantitative – This is generally numerical data, E.g. Statistics. This kind of data is collected using questionnaires and surveys

* Qualitative – This kind of data where people give ideas, this research is used most significantly for investigation purposes, E.g. Why people are buying a certain product

I previously said I would be using a method of sampling, but first I would like to explain what sampling is. Basic sampling is a representation of a large body by a smaller body. The advantages of this are that the company saves much more time, and it gives them the opinions of their customers. However the downside is this, not all people may have the same opinion, this therefore means that they won’t get their say as to how the company produce its goods. It is important with sampling that enough people are chosen to give confidence in the result. A sample size can be as low as 100 but obviously the more people asked, the more accurate the result obtained is.

Here are the various different methods of Sampling:

* Stratified Samples – Choosing people from a specific group in the population.

* Quota Samples – Selected on the basis of the characteristics of customer profile.

* Cluster Samples – People are chosen from a specific area.

* Systematic Samples – Every Nth person within a group is chosen.

* Random Samples – Everyone has an equal chance of being picked.

Of course, it is possible for an incorrect result from a questionnaire, this is could happen because of the wrong type of sample chosen or even because of poor facts and statistics. When compiling a question you should always…

* Show clarity and purpose to your enquiry

* Show understanding to all sides of community; try to word questions so that everyone can understand.

* Avoid misleading questions.

* Good presentation and structure.

The questionnaire is going to be made up of around 8 or 9 questions, and in it, it will include questions such as,

“Where do you buy your water?”

“How much do you spend on water per week?”

“Is water your preferred drink now?”

“Is drinking water really healthy?”

Of course these are only example questions and I would expect to have varied them a little in my final questionnaire.

To start of compiling my questionnaire I firstly drew up a rough page of ideas then begun work on Microsoft Pin point. Here I was able to lay down a question, and provide boxes for the reader to tick. I decided that a basic lay out was needed, so I ended up having a page with a border around, inset was the text of questions asking pupils about bottled water. The questionnaire had various different changes to be made, mostly because of the amount of space to provide for pupils to give their own ideas. Also the font was changed quite a bit from the original. After all the nit picking my questionnaire was finally ready, the next stage was the sampling. I decided that random selection was the fairest method so I randomly picked out two boys, and two girls from each class and stopped when I had reached a desired total. I handed them out to the selected people. Each of them filled out the questionnaire and gave it back to me. I then programmed all of my results into the computer.

In order to get a clear view of each opinion I decided that graphs, pie charts would be the fastest and easiest way to access the information. All graphs and a copy of the questionnaire are printed on separate pages. There are 3 graphs in total, the first graph is outlining the most popular flavour of water, graph two is outlining the most popular brand of water and the final graph is showing the results for how often water is drunk/purchased by the pupils of Castlederg high school. Overall there were no surprising results.

GRAPH ONE – This graph showed that the most popular water was “flavoured water”. It had around 43% of the vote; still water was second with 30% with sparkling water third on 27%.

GRAPH TWO – This showed river rock as the most popular brand with almost half of the votes.

GRAPH THREE – This showed the popularity of water when everyone said they drink water often or at least on a regular basis, no one said that they never drunk water.

Secondary research is data, which is published, either on hard copy or through an electronic source, which has been picked up by others for their own use. This kind of data must be taken with care, after all it is data, which has been used for a different purpose, and this may have affected the data analysis and presentation. When using secondary research companies should find out…

* Whether source is accurate

* Trustworthy

There are various kinds of secondary research, desk research is a quicker method than field research, and it is also less expensive. However the problem is that findings are not as accurate as field research. Secondary research can be obtained from internal or external sources.

The different forms of external research are:

* Government statistics

* Media

* The internet

* Market research companies such as key note, retail business

* Market surveys

The different forms of internal are:

* Databases of computers

* Sales invoices

* Complaint letters

* Sales information

* Financial information

External is probably a more important method of research as it doesn’t affect the companies’ performance to its competitors, but internal offers a snapshot view of the company as a whole and is information that is already held within the company. I have used external, mainly because of Internet research. Through external research I was able to go through different companies websites and find out information on their water. Classic water is used within our school and although it is a popular seller, the pupils decided that river rock was their favourite bottled water

In conclusion I feel that my research into bottled as went well, and have answered the questions I initially asked, why has it become more popular? The reason being pupils of my school think it is a healthy way of living. My results have shown many things, in particular the popularity of river rock within our school. Also showing that flavoured water is most preferred kind of bottled war.

Need help on conclusion, don’t know what to write!

Investigation Brief

Last night Mrs Leaf made two loaves of bread, she left them rising for half an hour and found that one rose a lot more than the other.

She used the following: –

0.5g yeast

10cm3 water

0.5g sucrose

10g flour

Investigate the fact that the leaves rose to different heights.

PLAN

AIM

In this investigation, I will be trying to find out what makes dough rise at different temperatures compared to my preliminary experiment in which I used only one temperature of 40 degrees. I will also try to make my measurements more accurate by taking more care thus leaving me with more accurate and reliable results.

APPARATUS

The apparatus that I will be using for the experiment will be the following:

10 test tubes

3 beakers

1 stirring glass rod

1 measuring cylinder

3 thermometers

Stop clock

Syringe

Yeast solution

Flour.

The apparatus is similar as used in the preliminary experiment except I will be using more test tubes.

I will be using two sets of test tubes at five different temperatures of room temperature, ice, 40 degrees, 60 degrees, and 80 degrees, with the same measurement of mixture.

For example: I will use 10ml of yeast to 10grams of flour in each test tube. I will then leave each tube at different temperatures of heated water.

I will be testing the best temperature for dough to rise three times at the selected temperature so that I know my results are accurate and to make sure it is a fair test.

The experiment consists of using dough. In the experiment I will be making the dough only by using yeast + sugar solution (liquid form) and flour. I will now explain the

science behind this experiment of what makes dough rise:

The yeast (a single celled fungus) becomes throthy once mixed with water and sugar. It is then mixed with flour and kneaded, thus enabling the flour to develop enough gluten to support the carbon dioxide made by the yeast. The dough is then left in a warm place for an hour. During this time the yeast cells multiply, this is fermentation. The carbon dioxide gas produced by the yeast forms pockets that makes the dough rise by doubling its size. This process is called proving. The loaf is then heated in an oven. The heat breaks down the yeast and evaporates the alcohol that is present in the yeast. It then cooks the dough leaving a risen loaf.

Changing the temperature (condition) might determine on how fast or slow the dough will rise and how much it will rise in mm compared to its original height before heating.

If the temperature of the water is higher then the dough will rise more and faster, if the temperature were lower the results would be the opposite. The temperature effects how fast or slow the dough will rise. I will be trying to find out if this hypothesis is true or not in this experiment.

METHOD

For the experiment we were given yeast solution and flour. The yeast mixture was made up of 0.5g yeast, 10cm cubed water and 0.5g sucrose. I was to use 10g of flour to put in each of the 12 boiling tubes and 10ml of the yeast mixture to make the test fair.

We weighed out the flour to exactly 10.0g on weighing scales and measured out exactly 10 ml of the yeast mixture. Our group mixed the two ingredients in each boiling tube with a glass rod putting the same measurements of yeast and flour in each tube. Then, with a yellow pen we marked on each boiling tube the starting height of the dough in mm before putting them in the water baths, we did this so that we could measure in mm at which height the dough rose. We also put labels showing the temperature of which water bath it was to be put in.

We controlled the water baths by putting a fixed temperature on each water bath. As soon as we put two boiling tubes in each water bath we started the stop clock. We used a stop clock to make sure that we did a fair test. We left the dough to rise for 30 minutes in the water baths.

As soon as 30 minutes passed we took the boiling tubes out and with the yellow pen marked where the dough had risen. I then, with a ruler measured how much the dough rose in mm and then put the results in my results table.

Compared with my preliminary experiment, I measured the mixture and flour more accurately and carefully. I did this by putting a mm scale on each of the boiling tubes to give me more accurate results .I also took the boiling tubes out of the water baths after exactly 30 minutes, I checked the times on the stop clock. As I have taken more care in this experiment I expect to have more reliable results, which will tell me if the hypothesis is true, or not.

Data Analysis

For my data presentation I showed a variety of graphs and tables. These included the class results, class average results, my own group results, my groups average results and a line graph showing my groups average results and the class average results. I have also included my preliminary results.

I decided to do average results for my group results and the class results in case the class results and mine were not exactly accurate. I showed these results in a line graph. It showed that the dough rose at its best at 40 degrees for the class average and my groups average, the result for class average at 40 degrees was 5.6cm and for my group average was 6.8 cm. It showed that after 40 degrees the dough did not rise a lot or not at all. The line on the graph started to decrease after 40 degrees. Before 40 degrees, the line graph shows that the dough is steadily rising but not so high. My group and class average line shows that there was not much difference in the results between the classes and mine. The curve of best fit indicates that the dough slowly rises and reaches its highest height at 40degrees and then slowly stops to rise so high.

The results will not all be the same for the class and my group as all the other groups may have not measured the mixtures as accurate or took the boiling tubes out at exactly 30 minutes and nor did our group so it is not surprising when seeing the class results that all results showed different figures. This is shown in my class results table that I drew on A4 paper in the data presentation. It shows that the general pattern was the same of the line graphs that I drew out.

I also did a table and graph showing the results for the experiment my group did. It shows the results for the two boiling tubes we put in each water bath in the table and line graph. They both show the dough mixture rose best at 40 degrees. I also drew a curve of best fit to clearly show that the height rises as it gets heated till it reaches its optimum height and show the line decreases as the dough denatures (as explained below).

The graphs and tables that I have drawn out, all clearly show that the dough rose at its best at 40 degrees compared to other temperatures and their results. This shows that temperature affects the way dough rises, whither it rises at all or rises quickly at high height. But why does temperature affect in which the dough rises? It is because when we raise the temperature the particles get more energy and move around more faster therefore leading to more collisions in a certain amount of time. The more collisions produce a reaction; in this case the reaction is making the dough rise. However the enzymes (a large protein molecule which acts as a biological catalyst) start to break down when the temperature gets too high as they stop working.

This is because when the enzymes get too hot it changes shape. This then means that the substrate (glucose) no longer fits into the enzymes active site. It is said that enzyme has ‘denatured’. It is when the yeast no longer consumes the glucose to produce alcohol and carbon dioxide, a process called fermentation. You can see in the graph that the line rises up to 40 degrees as it is heating up until it reaches it optimum height and temperature when the enzymes work at its best. The line then starts to go down which shows that the enzymes have started to denature so the dough does not rise any higher in height than the optimum height.

The diagram below shows the enzyme and substrate before it gets denatured, and another diagram showing the enzyme and substrate denatured:

Evaluation

I think that for this investigation, there was a good range of temperatures as we got good results that enabled me to work through this investigation and to find out at which temperature the dough rose best. I think that doing two repeats for each water bath helped in enabling me to do average results and to see whither my results were reliable by comparing the two results with each other.

However there were a few problems in doing the experiment but did not reflect the results and analysing them dramatically. The first problem that I encountered was stirring the mixture. This was because after stirring the glass rod picked up some of the dough mixture so there was not exactly 10ml of yeast mixture and 10g of flour in each boiling tube. Secondly there was the problem in measuring the exact height of the dough when using the yellow pen, as it was hard to see through the dough. Thirdly the timing was not always exact because at times we left for over 30 minutes but only slightly e.g. we left for an extra 20-30 seconds or a minute accidentally. This could have a slight effect on the results because within that 20-30 second period the dough could have rose a little bit more than it would have been at exactly 30 minutes.

Those were the only problems that I encountered with in the experiment. My data does not show any errors. I know this as looking at the class average and my average on the line graph, it shows that the results are similar and there are no points that go out of proportion. However in the class results table there was one group the had negative results meaning that the height of the dough decreased after it was heated at the temperatures of 20 degrees and 80 degrees. The group had probably made mistakes in doing the experiment so that’s why the results were negative compared to the other group’s results.

If I were to do this experiment again I would improve on the mistakes that I have just mentioned and I would do 3trys instead of 2trys so that I am sure of my results and therefore would no that they are reliable. I would also try different temperature of maybe 30 degrees or/and 50 degrees as both degrees are in between the optimum temperature of 40 degrees so either might also be the optimum height and temperature for dough.

Preparation of Primary Standard and Acid Base Titration

The purpose of this experiment is to determine the concentration of odium hydroxide solution by titration against the primary standard, ethanedioic acid-2-water.

Introduction

Titration is a method of analysis that to determine the precise endpoint of a reaction. In a titration, solution was run from a burette into a definite volume of another solution in a conical flask. 1 or 2 drops od suitable indicator (phenolphthalein) is added to indicate the end point. The end point is the stage at which two solutions have just reacted completely. The accuracy of titration results depends very much on the correct detection of the end point.

Chemicals:

Ethanedioic acid-2-water crystals

Dilute sodium hydroxide solution

Phenolphthalein

Deionized water

Apparatus:

Beakers (100 cm3 ) x 4

Conical flasks (250 cm3 )

Pipette (25.0 cm3 )

Pipette filler

Burette (50.0 cm3 )

Stand and clamp

Volumetric flask (250.0cm3 )

Wash bottle

White tile

Glass rod

Weighing bottle

Electronic balance

Stopper

Chemical Reaction involved:

Procedures:

1. Clean all the glassware involved in this experiment (e.g. burette, pipette, conical flasks, weighing bottle, volumetric flask, etc.) with deionized water as directed by the teacher.

2. Weigh by difference to collect the required mass (2 to 2.5g) of ethanedioic acid-2-water crystals.

3. Pour the weighed crystals into a dry clean 100 cm� beaker and add deinoized water to dissolve the crystals.

4. Wash the weighing bottle with deionized water and pour the rinse into the beaker also to ensure no loss of acid crystals.

5. Pour the content of the beaker in a clean 250 cm� volumetric flask and add up to the mark with deionized water.

6. Stopper and shake the flask thoroughly.

7. Wash a pipette first with distilled water and then with a small amount of sodium hydroxide solution given

8. Transfer 25.0 cm� of sodium hydroxide in a clean conical flask using the pipette and pipette filler.

9. Add 2 drops of phenolphthalein indicator to the conical flask. Phenolphthalein gives a red color in the alkali solution.

10. Wash a burette with distilled water and then with a small amount of the ethanedioic acid solution.

11. Close the stopcock. Transfer 50.0 cm� of standard ethanedioic acid solution in a clean burette.

12. Clamp the burette vertically in a stand.

13. Open the stopcock for a few seconds so as to fill the tip of the burette with sodium carbonate.

14. Record the initial burette reading.

15. Start the titration by opening the stopcock of the burette. During the process, gently swirls the conical flask continuously to mix the two solutions. At the time when the color changes from red to colorless, close the stopcock immediately.

16. Record the final burette reading.

17. Calculate the volume of ethanedioic acid required to neutralize 25.0 cm� of sodium hydroxide.

18. Empty the conical flask and wash it with water.

19. Repeat the steps 3 or4 times until the results from each trial are agreed.

20. After the experiment, rinse all the glassware used with tap water and put them back to their original positions.

Results:

Titration Results

Titration

1st (Trial)

2nd

3rd

4th

Initial reading (cm�)

0.00

17.00

0.00

15.50

Finial reading (cm�)

15.8

32.20

15.50

31.00

Volume of titrant (cm�)

15.8

15.20

15.50

15.50

Average volume of ethanedioic acid reacted:

(15.20+15.50+15.50)/3 cm�

= 15.40 cm�

That mean15.40 cm� of ethanedioic acid is used to titrate 25.0 cm� of sodium hydroxide.

Discussion:

1. Mass of ethanedioic acid crystals used:

= 2.50 g

No. of moles of ethanedioic acid:

= 2.50/ (12.0×2+1.0×2+16.0×4)

= 0.02778 mole

Molarity of ethanedioic acid

= 0.02778/0.250

= 0.11112M

No. of mole needed:

= 0.11112 x (15.40/1000) = 1.711 x 10��

Equation:

Mole ratio of C2H2O4 to NaOH is 1:2.

No. of moles of NaOH needed:

= 1.711 x 10�� x 2

= 3.422 x 10��

No. of mole = molarity x volume

3.422 x 10�� = molarity x (2.5 / 100)

molarity = 0.137M

Concentration of the sodium hydroxide is 0.14 mol/dm��

2. 3 requirements of a primary standard solution:

i) The solid dissolved in water should be pure.

ii) The molarity of the solution should be known.

iii) The solid must be dissolved completely.

3i) There is no effect if the burette is not rinsed with ethanedioic acid solution since we know the no. of mole of the ethanedioic acid and the volume of sodium hydroxide.

3ii) If the tip of the burette is not filled before titration begins, more acid is needed to titrate sodium hydroxide. By the mole ratio, since the no. of mole of sodium hydroxide is larger, and thus the molarity of calculated will be larger.

3iii) If the conical flask contains some distilled water before the addition of sodium hydroxide, the molarity calculated will be larger since the volume of sodium hydroxide increase.

4. It is advisable not to use sodium hydroxide solution to fill the burette in this experiment because alkaline will corrode glass and burette is expensive, therefore sodium hydroxide is usually placed in the conical flask.

Improvements:

1. Graduation mark of pipette and volumetric flask should be accurately reached.

2. Just add enough indicator solutions, not too much.

3. Ensure stopcock has been closed before filling.

4. Remember to fill up the tip of burette.

5. Fix the filled burette perfectly vertical.

6. Stopcock should be controlled by thumb, first and second fingers of left hand

7. Swing the flask gently and continuously.

8. Slow down the adding of solution when end-point is close.

9. There should be no hanging drop of solution on the tip.

10. The spectula should be put back the original place, to prevent the ethanedioic acid-2-water be contaminated.

Conclusion:

The concentration of sodium hydroxide calculated is 0.137M (correct to 3 sig. fig.)

Phi Function

My investigation is about the Phi Function ?. I am investigating the different ways on how to find the Phi Functions of different numbers and finding easier ways of finding the Phi Functions of large numbers. I will go through four parts for this coursework. I will start from the simplest cases of numbers and will go to more complicated. For any positive integer n, the Phi function ?(n) is defined as the number of positive integers less than n which have no factor (other than 1) in common (are co-prime) with n:

So ?(12)=4, because the positive integers less than 10 which have no factors other than 1, in common with 12 are 1, 5, 7, 11 i.e. 4 of them. These four numbers are not factors of 12.

Also ?(6)=2, because the positive integers less than 6 which have no factors other than 1, in common with 6 are 1, 5 i.e. 2 of them. These two numbers are not factors of two.

For the first part I will find the Phi Functions of many simple numbers and I will try to find a pattern on the Phi Functions of different types of numbers e.g. Odd numbers, even numbers, prime numbers, squared numbers, triangular numbers and so on. I will start from the numbers I obtained from the coursework sheet.

1. ?(3)=1, 2. Two of the numbers are not the factors of 3.

So ?(3)=2

2. ?(8)=1, 2, 3, 4, 5, 6, 7. Four of the numbers are not factors of 8.

So ?(8)=4

3. ?(11)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Ten of the numbers are not factors of 11.

So ?(11)=10

4. ?(24)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23.

I will now obtain the Phi Functions from numbers of my own choice and I will try to find patterns, if any. I will try to find patterns from even numbers, odd numbers, prime numbers, squared numbers, and triangular numbers.

EVEN NUMBERS:

?(2)=1

?(4)=2

?(6)=2

?(8)=4

?(10)=4

?(12)=4

?(14)=6

?(16)=8

?(18)=6

?(20)=8

ODD NUMBERS:

?(3)=2

?(5)=4

?(7)=6

?(9)=6

?(11)=10

?(13)=12

?(15)=8

?(17)=16

?(19)=18

?(21)=12

PRIME NUMBERS:

?(2)=1

?(3)=2

?(5)=4

?(7)=6

?(11)=10

?(13)=12

?(17)=16

?(19)=18

?(23)=22

?(29)=28

SQUARED NUMBERS:

?(4)=2

?(9)=6

?(16)=8

?(25)=20

?(36)=12

?(49)=42

?(64)=32

?(81)=54

?(100)=20

TRIANGULAR NUMBERS:

?(3)=2

?(6)=2

?(10)=4

?(15)=8

TRIANGULAR NUMBERS CONT’D:

?(21)=12

?(28)=12

?(36)=14

?(45)=24

?(55)=40

?(66)=20

I couldn’t find any patterns from any of the types of the numbers. But I have found out that there is an easier way of finding the Phi Functions of prime numbers. I have found a formula for finding the Phi Function for prime numbers, which is “P-1”. “P” stands for Prime number.

I the second part I am going to check whether ?(n�m)=?(n)�?(m) or ?(n�m)=?(n)�?(m). I am going to check the first two from the sheet.

1. ?(7�4)=?(7)�?(4)

?(28)=12. There are 12 numbers which are not factors of 28.

?(7)=6. There are 6 numbers which are not factors of 7.

?(4)=2. There are 2 numbers which are not factors of

4.

So ?(7�4)=?(7)�?(4)

2. ?(6�4)=?(6)�(4)

?(24)=8

?(6)=2

?(4)=2

So ?(6�4)=?(6)�?(4)

I am now going to check whether or not ?(nxm)=?(n)x?(m) for at least two separate choices of my own of n and m.

1. ?(2�8)=?(2)�?(8)

?(16)=8. There are 8 positive numbers less than 16 which are not factors of 16.

?(2)=1. There is 1 positive number less than 2 which is not a factor of 2.

?(8)=4. There are 4 positive numbers less than 8 which are not factors of 8.

So ?(2�8)=?(2)�?(8)

2. ?(7�3)=?(7)�?(3)

?(21)=12

?(7)=6

?(3)=2

So ?(7�3)=?(7)�?(3)

For the third part I am investigating why ?(n�m)=?(n)�?(m) whilst in other cases this is not so. I have found out that it depends on the types of numbers used as n and m. Now I will check whether or not ?(n�m)=?(n)�?(m) for different types of choices of n and m.

1. Even numbers of n and m.

?(6�4)=?(6)�?(4)

?(24)=8. There are 8 positive numbers less than 24 which are not factors of 24.

?(6)=2. There are 2 positive numbers less than 6 which are not factors of 6.

?(4)=2. There are 2 positive numbers less than 4 which are not factors of 4.

So ?(6�4)=?(6)�?(4)

2. Odd numbers of n and m.

?(9�9)=?(9)�?(9)

?(81)=54

?(9)=6

?(9)=6

So ?(9�9)=?(9)�?(9)

3. Co-prime numbers of n and m.

?(5×3)=?(5)x?(3)

?(15)=9. There are 9 positive numbers less than 15 which are not factors of 15.

?(5)=4. There are 4 positive numbers less than 5 which are not factors of 5.

?(3)=2. There are 2 positive numbers less than 2 which are not factors of 3.

So ?(5�3)=?(5)�?(3)

4. n(odd) and m(even).

?(9×4)=?(9)x?(4)

?(36)=14

?(9)=6

?(4)=2

So ?(9�4)=?(9)�?(4)

5. n(prime) and m(even).

?(7×4)=?(7)x?(4)

?(28)=12. There are 12 positive numbers which are not factors of 28.

?(7)=6. There are 6 positive integers which are not factors of 7.

?(4)=2. There are 2 positive integers which are not factors of 4.

So ?(7�4)=?(7)�?(4)

I have found out that there are two ways of having ?(nxm)=?(n)x?(m). The first way is that n and m should be co-prime. The other way is that either n or m should be prime and even. The rest of the number types result ?(n�m)=?(n)�?(m).

I have to investigate “?(Pn Q )” if P and Q are prime in the fourth part. I have found two formulas for finding the Phi Functions of large numbers. One of the formulas is “(Pn��)(P-1)”. The other formula is “Pn(1-1/P)”.

FIRST FORMULA:(Pn��)(P-1)

To find ?(81):

1. Firstly the 81 has to be split up into prime numbers.

81

3

27

3

9

3

3

3

1

81 converted into prime numbers=3

So the formula can now be used to find ?(81):

2. (3 ��)(3-1)=

3. 3��2

4. 54

So ?(81)=54

SECOND FORMULA: Pn(1-1/P)

To find ?(81):

1. Firstly 81 has to be split into prime numbers.

81

3

27

3

9

3

3

3

1

81 converted into prime numbers=3

So the formula can now be used to find ?(81):

2. 3 (1-1/3)=

3. 81�2/3=

4. 54

So ?(81)=54

The two formulas give out the same results for the ?(81). The “P” on the formulas stands for prime number. The formulas are faster at finding the Phi rather than writing all the numbers out and cancelling or circling them. There are four steps to work out the Phi’s from the formulas and that is why I have numbered my working out.

To investigate the factors that affect the amount energy produced

To investigate the factors that affect the amount energy produced in neutralisation reactions. The Aim of this investigation is to see how the dependant variable, the heat realised as a result of neutralisation reaction changes as one independent variable is changed, and to find why these changes occur. Only one variable will be changed. This is because if more than one is changed as well, we will not know which factor is responsible for the change. The variables are specified below, along with the one that I have decided to vary.

Variables

The following variables can be controlled during the experiment and will be the ones we can change in the investigation. The one that I have chosen is listed below. These variables are called independent variables, and will allow us to assess and investigate the effect on the heat released by neutralisation reactions.

1. The concentration of the acid or the alkali in the reaction could be decided to be varied (I have used the term alkali rather than base, because the substance will already be dissolved in water which is the definition of an alkali). To carry this out, one would have to obtain acids or bases or both of varied concentration, by obtaining a fairly strong concentration, and then diluting it down to get varied concentrations.

This procedure would be time consuming, and there would certainly be room for much error, as the concentration may not be measured out correctly, leading to inaccuracy. A general trend that would probably be seen is that, as the concentration goes up, so does the heat released by neutralisation. This is because there are more ions in a solution of a higher concentration. I have listed this variable as one, but it is really two different variables: one can either vary the concentration of the acid or that of the base.

2.The effect caused by the volume of the reactants could also be investigated. To do this, one would merely have to repeat the procedure, but using different volumes of the reactants each time. This procedure would be simple and safe, but if the experiment is wished to be very accurate, you would have to use a pipette, which proves to be time consuming. A pattern that would be visible when the different volumes of acid and alkali are mixed is that, as the volume rises, the heat of neutralisation too would go up, because there are again more ions in the volume to be neutralised. The conditions for this theory are as follows: the different volumes must be of the same concentration, if they are not of the same concentration the results would not show the correct pattern as two variables would be being altered.

3. If desired, it is also possible to vary the strength of the acid and/or base. By this, it is meant that a weak acid could be used, like Ethanoic acid, with a weak, and then strong alkali; the results could be compared to that which occurs when a strong acid is used with either alkali. The limitations of varying this factor are as follows: there is only one link between a strong and a weak acid, which is merely a scale called the pH scale. If this factor were investigated, we would obviously find that the combination of a strong acid and alkali would reproduce the highest energy rise. This is because the strong acids and alkalis dissociate to a higher degree, they split up completely into their composite ions. In weak acids, the degree of ionisation is less, and as a result of this, the number of ions in the solution is less, which prevents complete neutralisation.

4. The type of acid or alkali could also be used as a variable. This process would be investigated by using different acids and alkalis, whether they are strong or weak. The results of each acid and alkali (strong and weak) would be compared. This variable could be put under the same category of varying the strength of the acid. This is because as different types of acids and alkalis are being used, the strength of those acids and alkalis would also be being varied in the process. The other factors of whether the acid is Monoprotic, Diprotic or Triprotic all come under this heading. Acids, which form one H+ ion from each acid molecule, are called Monoprotic. Acids which form two are called Diprotic. Acids, which form three, are called Triprotic.

5. The last variable that could be used in the investigation is altering pressure of the acid and alkali. To do this, you would have to have an expensive, impractical piece of apparatus that would allow the pressure to be varied. The practical would be very difficult to perform and would be impractical. If you decided to use this procedure, you must also take safety precautions, because there is potential for the pressure container to either implode or explode. I believe that if the pressure were raised, there would not be much of a difference to the heat of neutralisation, unless it was raised fairly high which again is a limitation to this procedure. Another problem is that the equipment needed for this variable is not easy to obtain, and therefore couldn’t be used.

Introduction:

Neutralisation reactions

Acids and alkalis are defined as:

An Acid:

A substance that dissolves in water, producing H+ ions as the only

positive ions.

An acid is a substance, which contains hydrogen, which may be replaced by a metal to form a salt.

Properties:

They change moist litmus paper from blue to red.

They are soluble in water.

They are electrolytes.

They also have a sour or sharp taste.

Cautions:

Some acids are poisonous

Many acids are corrosive and thus dangerous. They burn flesh.

Acids as proton donors:

Acids produce hydrogen ions as the only positive ion. For example when hydrogen chloride dissolves in water the following process occurs.

HCL(aq) –> H+ (aq) + CL- (aq)

The hydrogen ion is sometimes called a proton. In water, the proton is combined with water as a result of the following process:

HCL(aq) + H2O(l) –>H3O+ (aq) + CL – (aq)

H30+ is known as a hydroxonium ion.

Hydrochloric acid has donated its protons to the water:

H+ (aq) + H2O (l) –> H30+ (aq)

All acids are proton donors.

Strong acids are fully ionized in water and are strong electrolytes. A strong acid produces a high concentration of H+ ions in a water solution. E.g. Hydrochloric acid. (HCl). Examples: sulphuric acid, hydrochloric acid and nitric acid. For instance, nitric acid:

HNO3 (aq) + H2O (l) –> H3O+ (aq) + NO3- (aq)

Weak acids are partially ionized in water and are weak electrolytes. Examples: Ethanoic acid. A weak acid: Produces a low concentration of H+ ions in a water solution. E.g. Ethanoic acid. (CH3CO2H)

Common strong acids include:

Hydrochloric acid (HCl)

Nitric acid (HNO3)

Sulphuric acid (H2SO4)

Common weak acids include:

Citric acid (H3C6H5O7)

Ethanoic acid (CH3COOH) (vinegar)

Alkalis and Bases

A soluble base is something which produces OH- ions in water.

A Base is a substance, which will react with an acid to form a salt.

A base is a proton acceptor.

An alkali is a base, which is soluble in water.

Properties:

They change litmus paper from red to blue.

They are electrolytes.

In addition many alkalis have a soapy feel.

All bases and alkalis, except ammonia, are metal oxides or metal hydroxides.

CAUTION: Many alkalis may be corrosive and poisonous. Example: sodium hydroxide is often called caustic soda. Caustic means ‘burning’.

A strong alkali:

Produces a high concentration of OH- ions in a water solution. Eg. Sodium hydroxide. (NaOH)

Strong alkalis are fully ionized in water and are strong electrolytes.

Weak alkali:

Weak alkalis are only partially ionized in water and are weak electrolytes. A weak alkali produces a low concentration of OH- ions in a water solution. E.g. Ammonia solution. (NH4OH)

Bases as proton acceptors- when a base reacts with an acid to form a salt, it accepts. Example: magnesium oxide reacts with sulphuric acid to form magnesium sulphate

MgO(s) + H2SO4 (aq) –> MgSO4 (aq) + H20 (l)

During this reaction the oxide ion, O2-, of the base accepts 2 protons H+ (O2- (s) + 2H+ (aq) –> H2O (l))

Common strong alkalis include:

Sodium hydroxide (NaOH)

Potassium hydroxide (KOH)

Common weak alkalis include:

Ammonium hydroxide (NH4OH)

Aluminium hydroxide (Al(OH)3)

Magnesium hydroxide (Mg(OH)2)

Hydroxide ions:

When alkalis dissolve in water an alkaline solution is formed. Alkaline solutions contain hydroxide ions. Example solid sodium hydroxide produces hydroxide ions when added to water.

NaOH(s) –> Na+(aq) + OH(aq)

These hydroxide ions accept protons to form water in the reactions between acids and alkalis.

H+ (aq) + OH- (aq) –> H2O (l)

Neutralization reactions

Acids react with bases to form salts.

Acid + Base –> salt + water

When aqueous solutions of an acid and a base are combined, a neutralisation reaction occurs. This reaction is characteristically very rapid and generally produces water and a salt. For a strong acid and a strong base in water, the neutralisation reaction is between the hydrogen and hydroxide ions dissolved in solution: H+ + OH- –> H2O

Neutralization is the reaction between an acid and a base in such quantities that only the salt + water are produced and no acid or base remain in the solution. When reacting both acid and alkali, both quantities must be as equal as possible, if a neutral solution is desired.

Strong acids and strong bases completely break up, or dissociate, into their constituent ions when they dissolve in water. In the case of hydrochloric acid, hydrogen ions, H+, and chloride ions, Cl-, are formed. In the case of sodium hydroxide, sodium ions, Na+, and hydroxide ions, OH-, are formed. The hydrogen and hydroxide ions readily unite to form water. If the number of hydrogen ions in the hydrochloric acid solution is equal to the number of hydroxide ions in the sodium hydroxide solution, complete neutralisation occurs when the two solutions are mixed.

Heat Involved in Chemical Reactions

The reaction of neutralisation is of course an exothermic reaction. This means that heat is given out during the chemical change that occurs. Along with all neutralisation reactions, all combustion reactions are exothermic, as they of course give out heat. The reactions, which are accompanied by a drop in temperature, are known as endothermic reactions; these reactions take in heat. When using a value of measure to the amount of heat given out, the end result is given a negative value for the change in energy. This may seem a bit odd due to the fact that it is clear that an exothermic reaction emits heat. The reason for giving exothermic reactions a negative ?H value is because the energy held by the substance has decreased, conversely, in an endothermic reaction, the energy of the actual substance has risen because the energy is held in the bonds.

The reason for heat being released from a reaction is because there are more bonds broken than are made, when bonds are broken, energy is taken in whereas the making of bonds leads to energy being produced. If the reaction is endothermic, there are obviously more bonds to be broken than have been made. It is also the case that stronger bonds take more energy to break than weaker bonds, and when stronger bonds are made, they release a greater amount of energy than when weaker bonds are created. Going by this, it is clear that every single reaction will have, to some degree, an energy change. Another factor discovered is that the amount of energy taken in by breaking bonds equals the amount of energy released through the creating of new bonds.

The amount of energy taken in or released can be expressed in kilojoules or joules, the SI unit for energy. To make the investigation fair I will express my values per mole. The energy changes that occur in reactions can be shown using energy level diagrams. In these diagrams, energy goes on the y-axis, and the x-axis is labelled as the reaction process, which shows the progress of the reaction. These diagrams do not show any numerical values, they are only used to show trends of energy changes in exothermic and endothermic reaction. The enthalpy diagrams are shown below, for both endothermic and exothermic reactions

The equation used to work out the energy transferred is ENERGY (KJ)= S.H.C X MASS (in g) X temperature change (in Kelvin)

Exothermic Reaction Endothermic Reaction

Products

Reactants

Products Reactants

Progress Of reaction Progress Of reaction

I have chosen to vary the type of the acid, for my investigation. I have decided that I am going to have a wide selection of different acids to investigate, but for the alkali, I am only going to have one weak and one strong one. I have chosen to vary the factor of type of acid, rather than volume or concentration (of either acid or alkali), because it would allow a simple and easy procedure. Another reason for choosing this variable is that it will allow a multitude of different combinations that will lead to clear-cut conclusions. The procedure for this variable is also somewhat less complex than most of the others. The acids that I will use for my experiments will be one molar values of sulphuric, hydrochloric, nitric, Ethanoic, methanoic and citric acid. I will use one molar Sodium hydroxide for the course of the investigation.

Preliminary experiment

Aim: to carry out the study on how the change in temperature of a neutralisation reaction is effect by the change in acid used in the reaction. I am going to use 6 different acids and 1 alkali for each experiment.

Prediction: I predict that the stronger acids will produce a higher temperature rise because, first of all, they have no bonds to be broken; in solution they exist as their component ions, completely dissociated. It is known that the breaking of bonds causes energy to be taken in, and when bonds are made, energy is given out. Another reason for a stronger acid producing a higher value for the heat of neutralisation is because it has more free H+ ions. I have deduced this because it is known that in a strong acid, all of the molecules are dissociated into their component ions. When the strong acid is used to neutralise the alkali, a more vigorous reaction would occur as a result of there being more H+ ions in the solution to neutralise the OH- ions in the alkali to give out more heat.

Apparatus:

2* 100 ml beakers

2* 500 ml beaker

2 measuring cylinders

Thermometer

Stirring rod

Method:

Collect the apparatus shown in the list above.

Measure 25 ML of acid.

Then place the acid in a measuring cylinder, to check if the volume is exactly 25 ml.

Then collect 25ml of alkali, and do the same as the acid except in a different measuring cylinder.

If the acid is Diprotic or Triprotic, and you are reacting it with a monoprotic alkali then you must use double or triple the volume of alkali in ratio to the amount of acid. This is to compensate for the extra H+ ions, which if not compensated for would result in an unfair experiment.

Measure the temperature for both acid and alkali and note the value down.

Then pour both into a beaker with the thermometer in the beaker as well.

Record the temperature rise. Perform this experiment for the rest of the acids.

Diagram for preliminary experiment

Results of preliminary experiment

Sulphuric

Sodium Hydroxide

20.0

20.0

32.0

12.0

Nitric

Sodium Hydroxide

21.0

20.0

28.0

7.5

Hydrochloric

Sodium Hydroxide

20.0

21.0

0.0

7.5

Ethanoic

Sodium Hydroxide

20.0

21.0

27.0

6.5

Methanoic

Sodium Hydroxide

21.0

19.0

27.5

7.5

Citric

Sodium Hydroxide

20.0

21.0

27.0

6.5

Acid

Alkali

Initial Acid Temp. �C

Initial Alkali Temp. �C

Final Temp. �C

Temp. Rise �C

Conclusion of preliminary results

The preliminary experiment performed was fairly well done, but there are several minor adjustments that could be made for when doing the real thing. Firstly, instead of using a beaker to mix the acid and alkali, a polystyrene cup could be used instead. This would stop energy being lost in the form of heat, to the surroundings. Also a lid would be placed on the polystyrene cup when reacting the acid and alkali to prevent further heat loss. I believe these are the only adjustments that need to be made for the real experiment. The prediction made was justified in the results processed, as the stronger acids produced a higher temperature rise due to the fact that, first of all, they have no bonds to be broken; in solution they exist as their component ions, completely dissociated. It is known that the breaking of bonds causes energy to be taken in, and when bonds are made, energy is given out. Another reason for a stronger acid producing a higher value for the heat of neutralisation is because it has more free H+ ions. I have deduced this because it is known that in a strong acid, all of the molecules are dissociated into their component ions. When the strong acid is used to neutralise the alkali, a more vigorous reaction would occur as a result of there being more H+ ions in the solution to neutralise the OH- ions in the alkali to give out more heat. For sulphuric acid there was a temperature change of 12*C, which was expected as it is a strong acid. However, for hydrochloric and nitric acid the temperature change was not as significant as expected. This could be due to wrong amounts of volume being mixed and more care will be taken when doing the actual experiment, to make sure equal amounts of volume are used. To avoid anomalous results the experiment could be repeated and I will do this in the real thing.

Actual Experiment

Aim: to carry out the study on how the change in temperature of a neutralisation reaction is effect by the change in acid used in the reaction. I am going to use 6 different acids and 1 alkali for the whole investigation, all with a concentration of one molar.

Prediction: I predict that the stronger acids will produce a higher temperature rise because, first of all, they have no bonds to be broken; in solution they exist as their component ions, completely dissociated. It is known that the breaking of bonds causes energy to be taken in, and when bonds are made, energy is given out. Another reason for a stronger acid producing a higher value for the heat of neutralisation is because it has more free H+ ions. I have deduced this because it is known that in a strong acid, all of the molecules are dissociated into their component ions. When the strong acid is used to neutralise the alkali, a more vigorous reaction would occur as a result of there being more H+ ions in the solution to neutralise the OH- ions in the alkali to give out more heat.

Apparatus:

2* 100 ml beakers

1* 500 ml beaker

2 measuring cylinders

2 Thermometers

Stirring rod

Polystyrene cup and lid

Method:

Collect the apparatus shown in the list above.

Measure 25 ML of acid.

Then place the acid in a measuring cylinder, to check if the volume is exactly 25 ml.

Then collect 25ml of alkali, and do the same as the acid except in a different measuring cylinder.

If the acid is Diprotic or Triprotic, and you are reacting it with a Monoprotic alkali then you must use double or triple the volume of alkali in ratio to the amount of acid. This is to compensate for the extra H+ ions, which if not compensated for would result in an unfair experiment.

Measure the temperature for both acid and alkali and note the value down.

Then pour both into a beaker with the thermometer in the polystyrene cup as well and seal the cup with the lid quickly and carefully.

Record the temperature rise and perform the experiment two more times. Perform this experiment for the rest of the acids, remembering to repeat it 3 times altogether for each acid.

Results for main experiment

Acid

Alkali

Initial Acid Temp. �C

Initial Alkali Temp. �C

Final Temp. �C

Average Acid Temp. �C

Average Alkali Temp. �C

Average Initial Temp. Of Acid + Alkali �C

Average of Final Temp. �C

Temp. Rise �C

1

2

3

1

2

3

1

2

3

Sulphuric

Sodium Hydroxide

21.0

20.0

20.0

22.0

20.0

21.0

32.0

31.0

32.0

20.3

21.0

20.7

31.7

11.0

Nitric

Sodium Hydroxide

21.0

21.0

21.0

21.0

21.0

21.0

28.0

28.0

28.0

21.0

21.0

21.0

28.0

7.0

Hydrochloric

Sodium Hydroxide

19.0

19.0

19.0

19.0

20.0

19.0

28.0

27.0

28.0

19.0

19.3

19.2

27.7

8.5

Ethanoic

Sodium Hydroxide

20.0

20.0

20.0

21.0

21.0

20.0

27.0

27.0

27.0

20.0

20.7

20.3

27.0

6.7

Methanoic

Sodium Hydroxide

19.0

18.0

19.0

20.0

20.0

21.0

27.0

26.0

25.0

18.7

20.3

19.5

26.0

6.5

Citric

Sodium Hydroxide

19.0

19.0

19.0

19.0

19.0

19.0

28.0

28.0

28.0

19.0

19.0

19.0

28.0

9.0

Analysis

As I predicted for the actual experiment’s results, the stronger acids reacted to give a bigger temperature than compared to the weaker acids. However, this was not the case for all of the acids used. Citric acid gave a very high reading of temperature change when it is known that it isn’t a very strong acid when compared with HCL and Nitric acid. But one should also take into account of the high volume citric acid used in reacting with the sodium hydroxide, as 75 cm(squared) of the citric acid were used to compensate for the fact that it is a Triprotic acid, as if the acid is Diprotic or Triprotic, and you are reacting it with a Monoprotic alkali (sodium hydroxide one molar in this case) then you must use double or triple the volume of alkali in ratio to the amount of acid. This is to compensate for the extra H+ ions, which if not compensated for would result in an unfair experiment.

The temperature change for sulphuric acid was recorded as being high, as was expected. As predicted the stronger acids gave higher temperature changes. This is due to the fact that stronger acids produce a higher temperature rise because, first of all, they have no bonds to be broken; in solution they exist as their component ions, completely dissociated. It is known that the breaking of bonds causes energy to be taken in, and when bonds are made, energy is given out. Another reason for a stronger acid producing a higher value for the heat of neutralisation is because it has more free H+ ions. I have deduced this because it is known that in a strong acid, all of the molecules are dissociated into their component ions. When the strong acid is used to neutralise the alkali, a more vigorous reaction would occur as a result of there being more H+ ions in the solution to neutralise the OH- ions in the alkali to give out more heat.

To aid my evaluation of my results I have calculated the enthalpy changes for each of the acids used. I will compare these results with the change in kelvin results.

Acid

Average temperature change/Kelvin

Specific Heat Capacity

Mass/grams (including alkali)

Energy change in Joules

Sulphuric

11.0

4.2

50.0

2310.0

Nitric

7.0

4.2

50.0

1470.0

Hydrochloric

8.5

4.2

50.0

1785.0

Ethanoic

6.7

4.2

50.0

1407.0

Methanoic

6.5

4.2

50.0

1365.0

Citric

9.0

4.2

100.0

3780.0

Change in Joules Graph

Change in Kelvin results graph

Further Analysis

The graphs for both change in Kelvin and Joules, have very similar patterns. The only difference between both is that Citric acid gives 3780.0 joules, which is 1470.0, more joules than sulphuric acid, which has a higher change in Kelvin than compared to citric acid.

The reason for there to be a higher amount of energy maybe due to the fact that a higher volume is used for the citric acid than sulphuric acid, which may link to the fact that a higher volume of acid gives a higher enthalpy change.

I believe my prediction was partially linked to the results recorded, mainly due to the fact that the weaker acids gave higher readings than expected, like Ethanoic and that the stronger acids gave lower than expected readings, like Hydrochloric acid. Overall I believe my results showed the trend that would be expected.

The Collision Theory

To help me explain the collision I am going to use information from “Chemistry Made Clear” by Gallagher and Ingram.

In order for a reaction to occur the particles must meet and the collision must occur with enough energy. In the reaction I am investigating, the making of magnesium sulphate, the magnesium atoms and hydrogen ions must collide with enough energy to successfully react.

Below are some diagrams to help show this:

The word equation for this reaction is below:

The symbol equation for this is below:

If there are many successful collisions i.e. the reaction goes quickly then a lot of hydrogen will be produced i.e. the rate of reaction is fast.

However if there isn’t enough energy to carry out the reaction then the hydrogen ion will bounce off and nothing will happen.

Acid Theory

When an acid dissolves in water the acid molecule splits up and hydrogen ions and anions are formed. Their quantity is different depending on the acid. If all the molecules of the acid split up completely then the acid will have a very acidic pH number. We know all acids contain hydrogen, but strong acids have a high concentration of hydrogen ions (H+) per dm3.

H2SO4 2H+ + SO42-

Due to the fact the hydrogen ions are in a solution then, as the particulate theory matter tells us, the ions will be moving randomly throughout the solution. Some will hit the metal. If the ions that hit the metal have enough energy then a chemical reaction will occur and the metal will lose electrons which become metal cations. These bond with the hydrogen ions and become hydrogen atoms. These atoms bond in pair to form molecules which effervesce and escape as gas.

Rates of Reaction

There are three factors that influence the speed of which magnesium reacts with sulphuric acid.

I will look at each of these in detail.

* The concentration of sulphuric acid.

If the concentration of sulphuric acid is increased then the speed of the reaction will be increased. This is because with an increase in concentration then there will be more hydrogen ions present. If there are more hydrogen ions then therefore the chances of a collision will be increased, and therefore the reaction will be faster.

* The surface area of the magnesium.

If a large block of magnesium is reacted with sulphuric acid then only a small amount of magnesium is actually exposed to the hydrogen ions. If the block of magnesium is broken up into small pieces then a larger area will be exposed. This increases the chances of collisions, and the more collisions they is the faster the rate of reaction.

* The temperature of sulphuric acid.

As mentioned earlier, particles need enough energy in the collision to react. This energy, also know as activation energy can be increased by increasing the temperature. So if the temperature of sulphuric acid is increased, then any collisions that occur will have the activation energy. This means the reaction will be faster.

Now for this course work I am required to investigate just one of these factors. I can forget about investigating the temperature of the sulphuric acid as to investigate it in detail may be a safety issue as heating an acid is unwise. It would also be difficult to investigate the surface area of magnesium as to do so I would need five different forms of magnesium and my school only stocks three of these.

So the easiest factor for me to investigate would be changing the concentration of sulphuric acid. This is quite easy as the actual experiment is quite straightforward and I can make my own dilutions of sulphuric acid.

Variables

The independent variable is what I am going to investigate; I will do so by changing the concentration of the acid.

The dependant variable is what I will be measuring, that is the time taken for all the magnesium to dissolve.

The controlled variables is what are going to stay the same to keep the investigation fair; i.e. the temperature of the acid, the temperature of the water, the total volume of acid solution, constant swirling and the size of the magnesium ribbon.

Prediction

My hypothesis is that the higher the concentration of sulphuric acid the faster the rate of reaction with the magnesium ribbon.

I have predicted this because:

All acids contain hydrogen, and hydrogen dissociates when it dissolves in water to produce ions. Sulphuric acid is one of the acids that dissociates completely to form a strong acid. Sulphuric acid will be very acidic on the pH scale and have a high concentration of hydrogen atoms as all its molecules will dissociate.

Its word equation is:

H2SO4 (aq) 2H+ (aq) + SO42-(aq)

As long as there is enough activation energy then a chemical reaction will take place between magnesium and sulphuric acid and magnesium sulphate and hydrogen will be produced.

The more successful collisions there are, the faster the rate of reaction.

So if the concentration of sulphuric acid is increased and as long as there is enough activation energy then the rate of reaction will be faster.

Apparatus:

2 burette holders

2 retort stands

2 waste beakers

1 stop clock

1 30cm ruler

1 pair scissors

1 small plastic filter funnel

1 piece sand paper

1 pair safety goggles

1 white tile

Chemicals

Sulphuric acid 100g/dm3

10cm length magnesium ribbon

Wash bottle of distilled water

Method

– Firstly I shall sand the piece of magnesium down to remove the oxide,

– Then I shall accurately divide my piece of magnesium into 10 1cm strips using scissors and a ruler.

– I shall then zero two burettes, one with water and the other with sulphuric acid.

– I shall then accurately add 25cm3 of sulphuric acid into my conical flask.

– To this I shall add a single 1cm strip of magnesium starting the stop clock as soon as it enters the flask.

– I shall listen and watch until the magnesium completely disappears then I shall stop the clock and record the time.

– I shall then repeat this using the same volume of sulphuric acid making sure to zero the burettes.

– I will then repeat the above from the third point until now using the values in my table.

– By dividing the two times I recorded for a piece of magnesium to dissolve by two I can find out the average time taken for each different concentration.

– By inverting my times as fractions (i.e. a time of 20 second which is 20 will become 1

1 20)

I can find out a decimal number which represents Rate of

Reactions-1.

Safety

During this experiment I shall wear safety goggles as Sulphuric acid is corrosive and can burn skin, and, as some may be taken up with the hydrogen ions as they escape as a gas (i.e. acid spray) it is only sensible to wear safety goggles.

Strategy for dealing with results

I shall display my results in a table like the one below:

Volume of Sulphuric Acid

cm3

Volume of Water cm3

Concentration of Sulphuric acid g per dm3

Time1

s

Time2

s

Average Time

s

(Rounded)

Rate of Reaction s-1

Rate of Reaction x 1000

s-1

25

0

25×100=100

25

17.28

14.54

16

1 =0.0625

16

62.5

23

2

23×100=92

25

18.78

17.78

18

1 =0.0555

18

55.5

18

7

18×100=72

25

29.69

28.84

29

1 =0.03448

29

34.48

16

9

16×100=64

25

40.84

39.90

40

1 =0.025

40

25

10

15

10×100=40

25

112

113

113

1 =0.008849

113

8.85

I will graphically demonstrate these results in a graph which I believe will look similar (if the times have been recorded accurately) to the one below:

I have predicted the positive correlation in the graph above as I believe that the rate of reaction is directly proportional to the concentration. I believe that as the concentration of acid doubles the rate should also roughly double. I believe this as when the concentration doubles there are double the number of hydrogen ions present, so there are double the number of ions to collide with. This would mean the reaction should occur in half the time, so the rate at which the reaction took place doubles.

I will make sure that the results I take are as accurate as they can be by using burettes instead of measuring cylinders as they are more accurate and have a control on how much liquid is poured.

I will make sure I read the meniscus at eye level as to make my reading more accurate. I will also use a white tile and put it behind the meniscus to help me see it clearer.

I will sand the magnesium as this will remove the oxide layer which forms and help the reaction to run smoother.

I will make sure to start the stop clock at the exact moment the magnesium drops into the sulphuric acid as the reaction will tart then. As well as this I will stop the clock as soon as the magnesium has completely dissolved as this is when the reaction is over.

Each different concentration shall have results taken twice to not only find an average time but to see if there has been an error. If an error has occurred I shall repeat the experiment a third time to find out which of my two previous times is incorrect.

Also I shall have to make sure the controlled variables I mentioned earlier are kept the same throughout, i.e.

The temperature of both liquid shall have to be kept the same, as will the total volume of the acid solution. I will have to make sure that all the pieces of magnesium ribbon are as close to one centimetre as I can possibly make them. The hardest variable I will have to keep the same shall be the constant swirling as it will be hard to keep it the same speed as the time before. All of the above have to be followed to make this experiment a fair test.

Obtaining Evidence

My Table of Results

Volume of Sulphuric Acid

cm3

Volume of Water cm3

Concentration of Sulphuric acid g per dm3

Time1

s

Time2

s

Average Time

s

(Rounded)

Rate of Reaction s-1

Rate of Reaction x 1000

s-1

25

0

25×100=100

25

17.28

14.54

16

1 =0.0625

16

62.5

23

2

23×100=92

25

18.78

17.78

18

1 =0.0555

18

55.5

18

7

18×100=72

25

29.69

28.84

29

1 =0.03448

29

34.48

16

9

16×100=64

25

40.84

39.90

40

1 =0.025

40

25

10

15

10×100=40

25

112

113

113

1 =0.008849

113

8.85

The above is my filled in table of results. All the times are my original times and as they seem quite consistent I felt no need to have to repeat an experiment.

I calculated the concentration by using the following calculation:

Volume of sulphuric acid x100=Concentration

Volume of water and acid i.e. 25

I calculated average time by using the following formula:

Time1 + Time2 = Average Time

2

I calculated Rate of Reaction by inverting the Average Time in Fraction and then I calculated its value as a decimal.

As you can see from my graph they appear to be quite accurate, but I shall go into more detail in the interpreting section.

Interpreting

I have drawn a graph to show the rate of reaction against the concentration of the solution. This graph is at the end of the interpreting section.

From the table you can see quite clearly that as the concentration of the solution decreases, the time take for the reaction to take place increases. This is what I predicted in my hypothesis.

Conclusion

I conclude that the rate of this reaction is directly proportional to the concentration of the acid which can be seen clearly from my graph. At a concentration of 40% the rate of reaction was 8.85s-1. The rate then quite steadily increased until, at a concentration of 100% the rate of reaction was 62.5s-1. These results strongly backup my conclusion. So, to increase the rate of reaction, the concentration of acid must be increased.

My original prediction was:

“The higher the concentration of sulphuric acid the faster the rate of reaction with the magnesium ribbon.”

And I believed, if I drew a graph, it would look like this:

I can now say that my original prediction was correct and my actual graph strongly matches the graph I made to show how I believe my results would look, i.e. there would be a strong trend showing that the rate of reaction is directly proportional to the concentration of the solution used.

Scientific Knowledge

Magnesium is high up the reactivity series (a list showing how reactive the elements are) and as it is higher up than Hydrogen it means magnesium is more reactive and can displace hydrogen in order to gain stability (a full outer shell of electrons). By placing a piece of magnesium in a aqueous solution of sulphuric acid, the magnesium displaces the hydrogen, which escapes as a gas .Meanwhile, the magnesium is forming into magnesium cations which bond with the sulphuric anions and together they form a salt. The hydrogen ions, move randomly throughout the solution, some of which, if there is enough energy collide and successfully react with the magnesium.

When the magnesium atoms loose atoms they become cations.

This is known as an oxidation reaction as hydrogen ions take the place of the displaced electrons and join together to form molecules.

As electrons are gained the below is a reduction reaction. The hydrogen molecules can escape as a gas which can be viewed and heard escaping from the solution.

The full equation of the experiment is

This experiment is an example of a redox reaction, due to the fact oxidation and reduction both taking place.

The magnesium is the reducing agent as it gives away its electrons quite easily.

The hydrogen ions are the oxidising agents in this experiment as they remove the electrons. Should the number of hydrogen ions increase, then more electrons can be removed and the reaction can occur faster. This can be observed when I used a 100% concentration of sulphuric acid, as there were a large number of hydrogen ions which increased the number of collisions with magnesium where electrons were lost, which increased the rate of reaction.

So, to conclude my experiment:

If the concentration of the solution is increased, and there is more sulphuric acid present then the rate of reaction with the piece of magnesium is increased.

Evaluating

My results

On my graph there is on point that is clearly off the line and two that are just barely on it. The one point that was off was my measurement for the rate of reaction taken for the solution with the concentration of 40gmd-3. It is this point however that I believe is anomalous, as the others are just on the line of best fit.

The point recorded for a concentration of 40gmd-3 has a rate of reaction faster than the expected trend. There are several reasons for this. The first is that possible I made a mistake in the measuring and added too much acid to the solution which would have given me a result that was too fast. There is also the chance I swirled the solution faster than I had with the other volumes, which would also explain a result off the trend line. Also there is the possibility too little magnesium was added; either I sanded the strip too much, and removed magnesium as well as the oxide layer, or I cut this particular piece too small, but either way it would explain a fast rate of reaction, or through no fault of my own this piece of magnesium was thinner than the rest, i.e. a fault of the supplier.

The last possibilities I can think of lie with the stop clock; either I stopped the stop clock too soon as perhaps the effervescence had dropped in noise level and the reaction was continuing when I thought it was over, or I started the experiment too late, while I attempted to drop the magnesium in, start swirling and start the clock all at once.

I will now evaluate the two points that are just on the line. To me they are not anomalous, but they do stand out so I will evaluate what could have gone wrong. For the point with a concentration of 92gmd-3 the above possibilities that I have for the point at concentration at 40gmd-3 apply.

However for the point at a concentration of 64gmd-3 there must be other reasons as it has a rate of reaction lower than the expected trend. It is possible that I didn’t swirl the solution as fast as I had done for the other concentrations. Also it is possible I made a mistake in the measuring of the magnesium and added a piece that was too big which would have meant the reaction would have gone on for longer. It’s also possible that I may have made a mistake in the volume of acid and added too little or I could have added too much water. Both of which could have given to a slower rate of reaction than expected. Then there is also the possibility that I may have not sanded the magnesium enough which could have left some of the oxide layer on, which would have slowed my reaction down. Finally it is possible that I stopped the clock too late.

Analysis of Apparatus

Measuring Apparatus

Burettes – I still think using the Burettes were a good choice as I could very accurately control the amount of acid and water I was adding.

Ruler- I felt this was a good choice as it enabled me to accurately measure out a 1cm strip of magnesium. However in combination with the scissors it was quite difficult to mark the length I needed to cut on the magnesium.

Stop Clock- The stop clock I felt was awkward, as I felt an extra pair of hands were needed to start the clock while I put the magnesium into the conical flask and began swirling. Unfortunately I cannot think of another piece of apparatus I could have used in its place, so it was and is still the best choice open to me.

Other Apparatus

Conical Flask- I feel a wider necked conical flask may have been better, as the piece of magnesium had a tendency to get stuck in the neck of the conical flask.

Scissors- I don’t really fell these were adequate in conjunction with a ruler to measure the strips of magnesium off with as I had to scratch the 1cm lengths on the piece of magnesium with these and then remove the ruler to cut the magnesium.

What would I do differently?

If I had to repeat this experiment there are several things I would do differently.

First of all I would use a magnetic stirring device instead of swirling the solution by hand which would remove the possibility of error.

Second of all, in conjunction with the magnetic stirrer I would use a beaker as there would be no need to worry about the acid spilling over the side with the magnetic stirrer. This would also mean the piece of magnesium couldn’t get stuck as the beaker has a very wide neck.

Finally, I would have used some fresh, powered magnesium. This would mean there would be no oxide layer to worry about. Also I would by able to measure it accurately with a scale which could eliminate error that I could have got from having to use a ruler and scissors.

Advice to the Medicine Company

I would advise the company to conduct some large scale tests to find which concentration of sulphuric acid would be safe, as to speed up their reactions I am sure they would use other factors which would increase the rate of reaction; which with a high concentration of acid could cause an unstable reaction.

An idea for the distribution of magnesium sulphate would be in the form of wipes which are easy to transport. The magnesium sulphate could be soaked into the wipes which would also prevent any magnesium sulphate being lost in the process because if any remained, it would be absorbed when more material is added.

Energy conversions

I have carried out a practical experiment based on energy conversions using a calorimeter and I am going to state the procedures of my practical experiment as well as calculating the amount of heat transfer.

I firstly started with weighting an empty calorimeter and recorded its mass.

Secondly I chose to weigh the same calorimeter with water (approximately 100g).

Thirdly I had to find the difference to get the mass of water.

Fourthly I measured the initial temperature of water in the calorimeter and recorded it.

Fifthly I decided to heat the water in the calorimeter to raise its temperature.

Sixthly I recorded the final temperature from the thermometer and the time taken.

Mass of empty calorimeter (g)

Mass of calorimeter with water (g)

Mass of water (g)

Initial temperature (oC)

Final temperature (oC)

Temperature difference (oC)

Time taken (sec)

Electricity to heat

23.2

123.2

100

18

31

13

302

Gas to Heat

72.5

173.5

101

19

32

13

44

Finally calculated the amount of heat transferred.

The table below shows the data that I collected from the practical experiment which I carried in the classroom:

Now that I have collected my data I am going to calculate the rate of heat transfer by using the following formulas:

Energy input = ITV

I = Current

T = Time

V = Volt

Energy output = mC?T

m = Mass

C = Specific heat capacity

?T = Temperature change

Electricity to heat

Energy input: I = 5 � T= 302 � V=7.5

= 11325

Energy input = 11325 (J)

Energy output: m = 100 � C = 4.2 � ?T = 13

= 5460

Energy output = 5460 (J)

Gas to Heat

The calorific value of British gas is 39.5kJ kG-1and I have used 320g of British gas.

Energy input: 320 � 39.5

= 12640 (J)

Energy input = 12640 (J)

Energy output: m = 101 � C = 4.2 � ?T = 13

= 5514.6

Energy output = 5514.6 (J)

Also I have been asked to calculate energy from Electricity to Kinetic of the following question:

Case study: A laboratory lifts uses an electrical motor to raise a carriage which weighs 35KN, through a vertical height of 10M at a steady speed, If the motor uses a current of 30amps at a potential difference of 450 and runs for 30sec.

Q1) How much work must it do?

A) The following equation is used for work done: W = FD

Work done = Force (N) � Distance (M)

As force is given as kilo Newton which means I would have to convert it

in order to get Newton by multiplying KN by 1000.

KN = 35 � 1000 = 35000 (N)

35000 (N) � 10 (M) = 350000 (J)

= 350 (KJ)

Q2) How much electrical energy does it use?

A) The following equation is used for electrical energy: E = ITV

Electrical energy = I (amps) � T (sec) � V (volts)

30 (amps) � 30 (sec) � 450 = 405000 (J)

= 405 (kJ)

As I have achieved to calculate the amount of energy given in and used I am able to calculate the efficiency of energy by using the stated formulas below:

Energy efficiency is the ratio between the useful output of an energy conversion machine and the input, in energy terms. The useful output may be electric power, mechanical work, or heat. Energy efficiency is not defined uniquely, but instead depends on the usefulness of the output. All or part of the heat produced from burning a fuel may become rejected waste heat if, for example, work is the desired output from a thermodynamic cycle.

Electricity to heat

Input energy = 11325 (J)

Output energy = 5460 (J)

5460 � 11325 � 100 = 48%

Energy efficiency = 48%

Gas to heat

Input energy = 12640 (J)

Output energy = 5514.6 (J)

5514.6 � 12640 � 100 = 44%

Energy efficiency = 44%

Electricity to Kinetic

Input energy = 405 (kJ)

Output energy = 350 (kJ)

350 � 405 � 100 = 86%

Energy efficiency = 86%

By looking at the stated results above I am going to make a judgment that energy from electricity to kinetic is more efficient where as electricity to heat is lesser efficient in this particular project and the reason for that is because when the gas is turned on from its source some of its energy loses in the air or losses energy where the Bunsen burner freshly turned on and it takes some time to heat it up while the energy is wasting.

As well as electricity to heat loses its energy when the electric device produces light and sound as waste energy. There are many pros and cons to consider when it comes to choosing between a gas heat transfer and electric heat transfer for any particular task. They both can be pretty efficient in their own way.

The down side of gas heat is in the name itself. The gas heat depends on fossil fuels to stay alive. The problem with that is most likely going to be the surge in prices when fossil fuels are in short supply. This can cause the bill to rise well over.

Newton’s Second Law

To confirm the relationship between the net force applied to a body and the body’s acceleration.

Background:

This experiment has two parts

> Part A: How does the acceleration of a trolley change as you change the weight of the total mass on a mass holder supported by a pulley?

> Part B: How does the acceleration of a trolley change as you change the weight of the total mass on the trolley itself?

When the trolley is released, it accelerates to the right due to the mass on the mass holder, passing through two light gates. A scalar timer connected to each light gate records the time taken for a card mounted on the glider to pass through the light gate The time recorded at the two light gates is used to calculate 2 different speeds of the glider, and hence the acceleration of the glider may be found.

Equipment: Trolley, 2 light gates, 2 scalar timers, masses, string, card, ruler, electric scale

Part A:

Data Collection:

Mass of ‘Weight’ (g)

Time (s)

Trial 1

Trial 2

Trial 3

Trial 4

LG1

LG2

LG1

LG2

LG1

LG2

LG1

LG2

30

0.84

0.44

0.87

0.45

0.85

0.44

0.86

0.44

40

0.62

0.33

0.64

0.32

0.63

0.32

0.62

0.32

50

0.52

0.28

0.53

0.27

0.53

0.27

0.53

0.27

60

0.45

0.24

0.46

0.24

0.46

0.24

0.46

0.24

70

0.42

0.22

0.42

0.22

0.41

0.21

0.41

0.21

80

0.39

0.20

0.39

0.20

0.39

0.20

0.39

0.20

90

0.37

0.19

0.36

0.19

0.36

0.19

0.36

0.19

100

0.34

0.17

0.34

0.18

0.34

0.17

0.34

0.18

110

0.32

0.17

0.32

0.17

0.32

0.17

0.32

0.17

120

0.30

0.16

0.30

0.16

0.30

0.16

0.30

0.16

Constants:

Mass of car 607.8 g

Distance between light gate 1(LG1) to light gate 2 (LG2): 30 cm

Card length: 17 cm

Data Processing:

To convert the masses to forces we must do the following. Since the measurements are taken in grams we need to divide by 1000. This makes our measurements change to kilograms. After that we need to multiply by 9.8 because 1 kg is 9.8 N.

Mass of Weight (g)

Force (N)

30

0.29

40

0.39

50

0.49

60

0.59

70

0.69

80

0.78

90

0.88

100

0.98

110

1.08

120

1.18

Since we have more than one trial for each mass, we need to find the average times for each mass.

Average Times:

Force (N)

LG1 Time (s)

LG2 Time (s)

0.29

0.86

0.44

0.39

0.63

0.32

0.49

0.53

0.27

0.59

0.46

0.24

0.69

0.42

0.22

0.78

0.39

0.20

0.88

0.36

0.19

0.98

0.34

0.18

1.08

0.32

0.17

1.18

0.30

0.16

If then to calculate the velocity we will take the length of the card (17 cm) which is d, and divide it by the average times.

Velocities:

Force (N)

LG1 Velocity (ms-1)

LG2 Velocity (ms-1)

0.29

0.20

0.38

0.39

0.27

0.53

0.49

0.32

0.62

0.59

0.37

0.71

0.69

0.41

0.79

0.78

0.44

0.85

0.88

0.47

0.89

0.98

0.50

0.97

1.08

0.53

1.00

1.18

0.57

1.06

To calculate the acceleration we will need to use the following equation for motion:

The next step is to make a (acceleration) the subject of the equation

This means that, where v is final velocity, u is initial velocity, and s is the distance traveled which is a fixed value of 30 cm (Distance between Light Gate 1 and Light Gate 2)

Acceleration:

Force (N)

Acceleration (ms-2)

0.29

0.18

0.39

0.34

0.49

0.48

0.59

0.61

0.69

0.76

0.78

0.89

0.88

0.97

0.98

1.16

1.08

1.20

1.18

1.35

There is a linear relationship in this graph; this shows that the net force is directly proportional to the acceleration.

Part B:

Data Collection:

Additional Mass of Cart (g)

Time (s)

Trial 1

Trial 2

Trial 3

Trial 4

LG1

LG2

LG1

LG2

LG1

LG2

LG1

LG2

0

0.39

0.20

0.39

0.20

0.40

0.20

0.40

0.20

50

0.42

0.21

0.43

0.22

0.42

0.21

0.43

0.22

100

0.46

0.23

0.46

0.23

0.45

0.23

0.45

0.23

150

0.49

0.25

0.50

0.25

0.49

0.25

0.49

0.25

200

0.51

0.26

0.50

0.25

0.52

0.27

0.51

0.26

250

0.54

0.28

0.55

0.28

0.54

0.28

0.55

0.28

300

0.58

0.29

0.57

0.29

0.59

0.29

0.57

0.29

350

0.61

0.31

0.61

0.31

0.61

0.31

0.61

0.31

400

0.66

0.34

0.65

0.33

0.66

0.35

0.65

0.34

450

0.79

0.39

0.78

0.39

0.80

0.40

0.78

0.38

Constants:

Mass of car: 407.7 g

Distance between light gate 1(LG1) and light gate 2 (LG2): 30 cm

Card length: 17 cm

Mass on Pulley: 50 g

Data Processing:

Mass of Cart:

To find the mass of the cart we need to add the initial mass of the cart to the additional mass added onto it. We also need to convert this number into kilograms.

Additional Mass on Cart (g)

Mass of Cart (g)

Mass of Cart (kg)

0

407.7

0.41

50

457.7

0.46

100

507.7

0.51

150

557.7

0.56

200

607.7

0.61

250

657.7

0.66

300

707.7

0.71

350

757.7

0.76

400

807.7

0.81

450

857.7

0.86

Since we have more than one trial for each mass, we need to find the average times for each mass.

Average Times:

Mass of Cart (kg)

LG1 Time (s)

LG2 Time (s)

0.41

0.40

0.20

0.46

0.43

0.22

0.51

0.46

0.23

0.56

0.49

0.25

0.61

0.51

0.26

0.66

0.55

0.28

0.71

0.58

0.29

0.76

0.61

0.31

0.81

0.66

0.34

0.86

0.79

0.39

If then to calculate the velocity we will take the length of the card (17 cm) which is d, and divide it by the average times.

Velocities:

Mass of Cart (kg)

LG1 Velocity (ms-1)

LG2 Velocity (ms-1)

0.41

0.43

0.85

0.46

0.40

0.79

0.51

0.37

0.74

0.56

0.35

0.68

0.61

0.33

0.65

0.66

0.31

0.61

0.71

0.29

0.59

0.76

0.28

0.55

0.81

0.26

0.50

0.86

0.22

0.44

To calculate the acceleration we will need to use the following equation for motion:

The next step is to make a (acceleration) the subject of the equation

This means that, where v is final velocity, u is initial velocity, and s is the distance traveled which is a fixed value of 30 cm (Distance between Light Gate 1 and Light Gate 2)

Acceleration:

Mass of Cart (kg)

Acceleration (ms-2)

0.41

0.90

0.46

0.77

0.51

0.68

0.56

0.57

0.61

0.52

0.66

0.46

0.71

0.44

0.76

0.37

0.81

0.30

0.86

0.24

Since we are getting a hyperbola shape in the graph, we will try to graph Mass of Cart vs. Acceleration-1 to get a linear relationship.

I think that the last two points are outliers and they have been subject to lots of error. This is why I will draw the graph again without including them in the trend line.

These graphs show that the mass of the cart is inversely or indirectly proportional to the acceleration. This means that the mass of the cart is directly proportional to the inverse of the acceleration.

Conclusion and Evaluation:

Figure 1

Figure 2

The aim of this lab is to confirm the relationship between the net force applied to a body and the body’s acceleration. The experiment has two parts: Part A’s aim is to see how the net force applied to a body affects the acceleration and Part B’s aim is to see what happens to the acceleration as you increase the mass of the cart.

Obviously, the higher the net force on a body, the larger its acceleration. This is the obvious thing we found out from part a. In part B we found out that as the mass of the cart increased, its acceleration decreased. This is where the errors come in.

This experiment was designed in a way the friction would be very minute or inexistent. The first set-up is using an air track; this allows the vehicle to move freely without friction due to a cushion of air being blown through the air track. The second set-up is using a friction compensated runway; this runway cancels out the effect of friction. However, our set-up included friction; this means that our results are distorted. The other thing is that all the values for the acceleration are lower than the actual value. This is because of Friction; all these calculations neglect the effect of friction. Friction makes a body move slower, slow moving bodies have a large inertia, so all the values of Mass are larger than the actual value.

Assume there are two bodies, body A and body B. They are traveling on different surfaces but are pushed with the same force. Body A has a smaller net force than body B; therefore you assume that Body A has a larger mass than Body B or Body B has a larger acceleration than Body A. What if I told you that this is not the case? Maybe Body A and Body B have the same mass, what would you think? It would be logical to think that there are other forces acting on the system, Friction. Body A is traveling on a wooden surface (friction is present), Body B is traveling on a slippery, frictionless surface (friction is not present). So Body A doesn’t have a larger mass than Body B, it’s just the surfaces that make us think so. This is the same here; these calculations assumed that friction was something else, which is why there are absurd numbers.

If then imagine that a tension force of 1 N is acting on a body which is accelerating a 5 meters per second squared. and and This means the mass of the body equals 0.2 kilograms. What is happening in our situation is that our acceleration is smaller than it should be due to the other forces acting on the cart, like friction and air resistance. For example, instead of 5 ms-2 it is 4ms-2. This causes for there to be an error in our calculations.

and and . This means the mass of the body is 0.25 kilograms- but it’s not. This value is larger than the real value, and this is exactly what is happening to us.

The same problem exists in Part B. The net force is always less than the real value, 0.49 N. This is due to the other forces acting on the body like friction and air resistance.

, in part B the net force is always constant and the mass and acceleration are always changing. Since the acceleration is going to be smaller than it should be (due to the other forces acting on the body) the net force will also be smaller than it should be.

The other problem with including friction comes in part B. When you add weights on the cart, you increase the force of gravity acting on the cart and the force of normal reaction acting on the cart (Figure 2). But, as you increase the mass of the cart you push the cart and make it come closer to the surface it is in contact with. This increases the force of friction and makes is increase as you increase the mass of the cart.

The other (smaller) errors in our lab were the following. Firstly, the way we released the weight that would pull the cart was different each time, there is no way of keeping that the same no matter how hard we try. Secondly, we drew a line that shows where we are supposed to release the cart; the cart will never be in the same position as it was the first time we released it. Finally, we did this lab in two sittings, this means the environment changed and we had to make some adjustments, this affected the accuracy of our lab.

To ensure the accuracy of our lab we did many things. Firstly, we measured the distance between the light gates more than one and between trials. Secondly, we made sure that the light gates were placed parallel to the edge of the counter. Also, we drew a line that would show us where we should release the cart. Next, we did four trials instead of three or one trial. We weighed the cart more than once and we measured the length of the card more than once as well. Finally, we tried to keep the lab as accurate as possible by doing the same jobs each time because everybody does their own job in their own way and if that way was kept constant than our results would be more accurate.

Next time, we will use a different set-up, preferably with an air-track and a very streamlined glider. We will take even more accurate measurements so that our data would be even more accurate than it is. We will make sure we are releasing the cart from the same place each time, even if it means we need to use a magnifying glass. We will also try to do this lab in one sitting. Finally, we will try to come up with a way in which we will release the cart so that it would be the same each time.

To investigate and understand the relationship between the kinetic energy and stopping distance of a toy car

I predict that the when the vehicle is tested at 10 books high there will be a greater amount of gravitational potential energy than at 5 books high, which will be converted into kinetic energy when the car is released; giving the car a longer stopping distance.

I think the cars at 10 books high will travel further. I think that the times taken for the cars to stop at both heights will be similar but in general the 5 books will take longer time to stop. I think that the heaviest car will travel the furthest distance.

Fair Test:

I will make my test fair by:

* Instead of trying to push the cars from their stationary positions, I will place them all onto the ramp and let them go, so that no cars have an unfair head start.

* Also I will make sure that the starting positions of all of the cars is the same and this will be done by marking a starting line on the ramps.

* By measuring the height and angle of 5 books and 10 books so that when continuing the experiment no car would have an advantage

* By using the same ramp to keep consistency

Method:

1. Collect equipment and assemble as shown in diagram.

2. Choose 5 books to put under ramp

3. Get a toy car and weigh it

4. Release car from the top of the ramp

5. Use a stopwatch to time the how long it takes the car to stop

6. Measure the length travelled by the toy car

7. Repeat Steps 3-5 twice more

8. Repeat Steps 3-6 with 10 books underneath the ramp

9. Repeat Steps 3-7 using another car

10. Repeat Steps 3-8 using a third car

11. Record results

12. Pack up

Results:

Length (metres)

Time (secs)

Weight (grams)

Red Car

5 books high

1st

1.96

4:84

34.5

2nd

1.83

4:97

3rd

1.88

4:40

10 books high

1st

2.95

4:75

2nd

2.87

4:54

3rd

2.06

4:60

Blue Car

10 books high

1st

2.15

3:65

40.3

2nd

2.17

3:62

3rd

2.30

4:28

5 books high

1st

1.77

4:65

2nd

1.71

4:81

3rd

1.52

4:79

Yellow Car

10 books high

1st

2.53

3:92

52.9

2nd

2.20

4:12

3rd

2.00

3:49

5 books high

1st

1.58

4:49

2nd

1.63

4:74

3rd

1.41

3:90

Observations

From this experiment I have observed that more the amount of gravitational potential energy there is then when converted into kinetic energy and the car is released there is a longer stopping distance.

The cars’ molecules have very weak bonds with other molecules. When the two surfaces come close together, the forces make them act as if they are slightly sticky, and this will also slow down the rolling movement by acting as a brake. Inertia is the thing that makes it difficult to make a big mass start to move or stop it once it is moving. If the car is stationary then you have to use a force to overcome its inertia and get it to move. If the car is moving then you have to use a force to overcome its inertia and stop it moving.

Conclusion:

I conclude that the relationship between the stopping distance and kinetic energy is that the more of the kinetic energy that there is, the longer the stopping distance.

The fastest car was the yellow car on all three attempts. The yellow car was also our heaviest car. The greater the mass, the larger the acceleration needed to slow it down. As the force of friction is not increasing all that much in proportion to the increased mass, the more mass an object has, the longer it will take to decelerate and the longer its braking distance will be.

There are 2 sorts of friction. These are static and dynamic friction in order for the cars to move they had to overcome both of these.

Investigating the effect of concentration on the rate of reaction between marble chips and Hydrochloric acid

In this experiment I will be investigating the reaction between different concentrations of acid and marble chips. I will be measuring the speed and rate of which gas is given off. I know that the gas given off will be carbon dioxide and that the reaction will be as follows:

Calcium Carbonate + Hydrochloric Acid –> Calcium Chloride + Water + Carbon Dioxide

Or

CaCO3 + 2HCl –> CaCl2 + H2O + CO2

On the left are the reactants. These are the chemicals that I will start off with. On the right are the products. These are what will be produced at the end of the experiment. Reactants and products are very rarely the same, as they react with one another to form other things. There will always be the same number of atoms in the products as there are in the reactants though, as none can be dragged in, or left out.

Variables

In this experiment I am going to change the concentration of Hydrochloric acid. I will use 2M Hydrochloric acid as 100% concentration. Here is a table to show all of the concentrations I am planning on using, and how much acid and water to put into the conical flask. It also states how strong the acid is in M.

Concentration (%)

Molar

Hydrochloric acid (ml)

Water (cm�)

100

2.0

20

0

90

1.8

18

2

80

1.6

16

4

70

1.4

14

6

60

1.2

12

8

50

1.0

10

10

I will then investigate the difference in speed of which the gas is given off, over a two minute period. I will take measurements every 10 seconds.

Preliminary Work

Before this experiment, I decided which size of marble chips to use from the sizes available. The different sizes of marbles chips had an average diameter of: large-1.5cm, medium-1cm, or small-0.5cm. After a short experiment which included putting 3 of each size, in turn, into a conical flask, with hydrochloric acid in. I then measured the amount of gas that came off using the same technique as I used in the final experiment. I then noticed that the small marble chips produced gas so slowly that it would have been hard to accurately measure the amount of carbon dioxide given off. The large marble chips, however, gave off carbon dioxide too fast. Therefore I decided to use the medium sized marble chips. This is because 3 large marble chips have a greater area than 3 small marble chips, so more hydrochloric acid can get to the large marble chips, so more particles collide, resulting in more gas being produced faster.

Prediction

I predict that the higher the concentration of acid, the faster gas will be given off.

I also predict that the gas will be given off twice as fast with 100% concentration than it will with 50% concentration, because their would be twice as many hydrochloric acid molecules to hit each other, so the molecules will collide twice as often.

Apparatus

In this investigation I will use:

A conical flask

A gas delivery tube

A margarine tub

A measuring cylinder

A clamp

A stopwatch

Marble chips

Hydrochloric acid

Water

Diagram

Method

– First, I will gather all of the equipment listed on the previous page.

– I will then make up my concentration of acid and put it into the conical flask.

– Next, I will fill the margarine tub between 1/2 and 3/4 full.

– I will then fill the measuring cylinder full of water, and tip it upside down in the margarine tub, ensuring that as little air as possible gets in.

– The measuring cylinder will then be secured upright using the clamp.

– Then, I will put 5 marble chips, of roughly equal surface area, into the acid.

– I will then put the bung of the gas delivery tube into the water in the margarine tub.

– When the gas begins to bubble out of the end of the tube, I will place it under the measuring cylinder to collect the gas.

– At the same time as the previous step, I will set the stopwatch going.

– I will then, note the amount of gas collected every 10 seconds.

– I will then repeat the experiment as many times as need be, using different concentrations of acid.

Measurements to be Taken

In this investigation I will be measuring the amount and rate that carbon dioxide is produced, when marble chips (calcium carbonate) are put into hydrochloric acid of varying strengths. I am going to test each concentration 3 times to make sure that the investigation is fair, and will then work out averages, rates of production, gradients etc.

I will measure the amount of carbon dioxide in cubic centimetres, and will take measurements every 10 seconds. I will be collecting the gas using a measuring cylinder that has been upturned in a margarine tub between 1/2 and 3/4 full of water.

Fair Tests

To make this investigation fair, I will be using the same amount of marble chips, and will ensure that they have roughly the same surface area. I will do this because if I used marble chips which had different surface areas then more acid would be able to reach the marble so more gas would be given off in less time.

I will also take measurements as soon as possible after the gas starts to comeout of the tube. This is because as the marble chips dissolve into the acid their surface area gets smaller, resulting in gas being given off slower.

I also tried to keep the temperature about the same throughout the experiment. This is because with more heat energy the particles will gain more energy, and therefore move faster. If the particles move faster, there will be more collisions and therefore the reaction will take place faster.

I also kept the time between measurements the same. If I had waited 12 seconds instead of 10 for one of the measurements, the amount of gas measured would not be a true measurement. Therefore, I used a stopwatch to measure time.

Safety

To help make this experiment safe I will wear goggles at all times during the experiment or whilst others are doing the experiment near me. This is so that if for any reason acid gets splashed at me, it will not go in my eyes. If acid did get in my eyes it could damage them.

I will also stand up whilst doing the experiment because if acid gets spilled, I will be able to get out of the way quickly so that it doesn’t get on me or my clothes. If I was sitting down it would be harder for me to stop the acid from getting onto me.

If I spill any acid onto me I will rinse the area under the tap, to get the acid off. I would need to do this because hydrochloric acid is corrosive and therefore would hurt or even damage me.

I will not run or mess about in the laboratory in case I spill acid, which would cause a safety hazard for the same reason as above.

The concentrations of acid I am using in this experiment are so weak that it doesn’t really matter too much if I do come into contact with it. I will still carry out the safety precautions however, so that there is no possible danger.

Section 2: Obtaining

Results Table

My results tables are at the end.

Section 3: Analysis

Analysis

From my results I can see that generally as the acid that I used got weaker, the less gas was produced. This is shown by the fact that with 100% concentration of acid I measured an average of 69.67cm� of carbon dioxide, the average gradually decreases apart from at 70% where it increases, but from then onwards goes steadily down.

In the 100% concentration test, where I used 2M Hydrochloric acid, the volume of gas had increased by about 6, 10 seconds after the last measurement. With the 90% concentration test, I have decided to ignore the first set of measurements because I feel that they are not accurate. Therefore I have remade the table as follows:

A table of results for 100% concentration of hydrochloric acid, excluding the first repetetion

Time (s)

Repetition 2 (cm�)

Repetition 3 (cm�)

Average amount of gas measured (cm�)

0

0

0

0.00

10

3

7

5.00

20

7

9

8.00

30

10

12

11.00

40

15

17

16.00

50

18

21

19.50

60

21

25

23.00

70

25

29

27.00

80

28

34

31.00

90

31

38

34.50

100

34

42

38.00

110

38

45

41.50

120

41

49

45.00

I can see from this that the amount of gas measured goes up by about 4 cm� every 10 seconds.

During the 80% test the gas produced increases by about 3cm� every 10 seconds.

In the 70% concentration test the gas produced increases by about 4cm� every 10 seconds, this is strange because, logically it should increase by less than the 80% concentration test. By surveying the other results and the patterns that have emerged from them I would estimate that it should have increased by 2cm� every 10 seconds instead of by 4cm�. I think that this has happened because the marble chips were probably larger. This is explained in more detail under the section entitled “Anomalous Results”

In the rest of the tests the rate that carbon dioxide is given off, seems to level off at about 3-4cm� every 10 seconds.

Conclusion

From this investigation I have learned that if an acid is concentrated it is more reactive. Being more reactive means that for instance, in this experiment, more carbon dioxide is produced and at a faster rate. If the acid were less reactive the reaction would take place slower, therefore less carbon dioxide would be produced overall, and at a slower rate.

Proving the Prediction

On the whole, my prediction was more or less right. As the acid got weaker, the amount of gas produced, decreased, apart from at the 70% concentration, where it instantaneously increased. This was probably caused by using marble chips that had a larger surface area to the ones I had been using before. This was due to the fact that I had to slightly adjust the size of the marble chips I was using, as there were no more marble chips that were the same size as the ones that I had been using previously. To get avoid this problem I could have used specially prepared marble chips, that had a fixed surface area, such as 1cm� cubes. This would have increased accuracy, but would also have increased the cost of the investigation by quite a lot.

I also predicted though that there would twice as much gas produced with 100% concentration than there would have been with 50% concentration. This was nearly right as the average volume of gas produced with 100% concentration was 69.67cm� and the volume of gas produced with 50% concentration was only 29.33cm�. Rounded up to the nearest whole number these work out at about 2.5 times more gas produced. I think that this is near enough to my prediction.

Section 4: Evaluation

Evaluation of Evidence

I do not think that all of my results are reliable. There seems to be major differences between some of the results, for instance the first repetition of the 90% concentration test, produced carbon dioxide at higher rate than that of the 100% concentration test. It also has a higher overall volume than the 100% concentration test and is double that of the other repetitions.

I also do not think that the 70% and 60% results are reliable because they both have a larger overall volume of gas produced than the 80% test.

Anomalous Results

During this investigation I did get some strange results. In the first version of the 90% concentration, overall more carbon dioxide was produced than in any of the 100% concentrations. This should not have happened because the acid was weaker so the reaction should have taken longer. The reaction took longer in all of the other repetitions, so why did it take longer in this case?

There are several reasons why this could have happened. These are: That the acid could have been stronger; or the surface area of the marble chips could have been bigger. The first of these is hard to justify, but could have happened if by accident more acid than was supposed to be put in was put in. This could have happened but it is impossible that the acid was more than 100% concentration, which it would have needed to be in order to produce more gas than the 100% concentration experiment. This only leaves the second explanation: That the surface area of the chips was larger.

This could have been for any reason, such as: The marble chips were larger than in the other tests; more marble chips than should have been put in, were put in; or maybe that one of the marble chips had a crack in. I would guess that more marble chips were put in, although it could have been because of any of the other reasons. From the fact that almost twice as much gas was collected in the first repetition, I would say that either 2 lots of marble chips were put in by accident, or the marble chips from the last test (the 3rd repetition of the 100% concentration test) were still in the conical flask.

To stop this happening again, if I did another investigation, would be to always check that the experiment has been set up correctly with the right quantities and concentrations of everything.

Also the average volumes of gas with the 70%, and 60% concentrations of acid, are higher than that of the 80% concentration test. This should not have happened because the acid should have been weaker. This was probably caused by using marble chips that had a larger surface area to the ones I had been using before. This meant that there was more marble for the acid to react with, and therefore produced more gas.

In order to stop this happening, if I did the experiment again I could use marble chips which all have exactly the same surface area. This would increase accuracy but unfortunately make the investigation to expensive to carry out, as the marble chips would probably have to be specially prepared.

Evaluation of Method

I think that my method worked quite well, although next time I would use marble chips with a set surface area. This would improve the accuracy of the investigation. Also I would check to make sure that the experiment was set up properly with the right amounts of the chemicals and substances that I was using. I would do this to ensure my results were accurate.

Is the Conclusion Valid?

I think that my conclusion is valid because my colleagues all had pretty much the same sort of results.

Results

All of the units in the following tables are in cm�, apart from the units in the “Time” columns which are in seconds