C5 Questions Warning (plenty of them)

Hello guys, I've done some C5 questions, I just need confirmation if they are right. I think most of them are correct, but for some I want to know if that's a good exam-styled answer.

Key: Question Answered

Quantitative Analysis

12. What are the advantages and disadvantages of the traffic light system?

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The advantages are that it tells you which food contains a low or medium or high amount of each food groups. It also gives an indication as to which food is healthy and which is unhealthy. The traffic light system also tells you the amount of 'fat', 'sugar' or 'salt' per serving and then tells you if it is a low, medium or high amount.
The disadvantages are that it does not tell you how much % of the GDA each serving contains - for example, it says there is 2g of saturated fat which is 'low', but it doesn't tell you that it is 10% of the GDA for saturated fat (g).

13. A tin of beans, A, contains 1.2g of sodium. Tin B contains 2.8g of salt. Which tin, A or B, contains more sodium?

Method 1: Tin A contains 1.2g of sodium.
Tin B contains 2.8g of salt.
Na in NaCl = 23 / (23+35.5) = 23 / 58.5 = 0.393 (3 s.f.)
So, for every 1g of salt, there is approximately 0.393g of sodium.
So, the amount of sodium in 2.8g of salt would be: 0.393 * 2.8 = 1.1004g of sodium.

Method 2: Tin A contains 1.2g of sodium.
Tin B contains 2.8g of salt.
58.5 / 23 = 2.543 (3 d.p)
There is 1g of sodium in every 2.543g of salt.
2.8 / 2.543 = 1.101 (3 d.p)
1.101 * 1 = 1.101g of sodium.

The actual answer I get is 1.1008547008547, so the 2nd method would give a more accurate answer.
Can someone confirm if that is the right answer?

15. Some food labels only give the amount of added salt. Explain why this is not an accurate indication of the total amount of sodium.

Some foods may contain other sodium compounds such as sodium bicarbonate which is different to salt but yet it contains sodium. Therefore, salt is not an accurate measurement of the amount of sodium present in a food.

Titrations

7. (b) Explain how to estimate the end point from a titration curve.
The end point could be estimated as you can identify it from an abrupt change in the pH on the titration curve and also from the largest vertical section of the curve.

8. Explain why for titrations a single indicator is better than a mixed indicator.

Single indicators only show one colour change which makes it easy for us to spot the end-point of the titration. Universal indicators constantly change colour at nearly every pH change which will make it very hard to see the end-point during the reaction.

10. Explain why it is important to use a pipette filler when filling a pipette with alkali.

It is used as an alternative to allow the pipette to suck up the alkali until it is filled with the required volume. This is a safer method to avoid any hazardous chemical reactions occuring in the body.

12. Which titrations should be used to calculate the average?

This is not the full question, but this is the only part I wanted confirmation on.
The titration results which are close together and have small differences should be used to work out the average.

13. Explain why phenolphthalein is a better indicator than a universal indicator.

Phenolphthalein is a single indicator so it will only have one colour change during the reaction which will allow us to spot the end point and the neutralisation of the acid/alkali easily.

14. 26.1 cm³ of hydrochloric acid react with 25.0 cm³ of 0.095 mol/dm³ sodium hydroxide solution. Calculate the concentration of the acid.

Moles in alkali = Volume in dm³ * concentration
= (25/1000) * 0.095
= 0.025 * 0.095
= 0.002375

1:1 ratio for the number of moles as 1 mole of HCl reacts with 1 mole of NaOH in the reaction.

Number of moles in alkali = Number of moles in acid
0.002375 = alkali concentration * volume in dm³
0.002375 = alkali concentration * (26.1/1000)
0.002375 = alkali concentration * 0.0261
0.002375 / 0.0261 = alkali concentration
alkali concentration = 0.091 mol/dm³ (2 s.f.)

Gas Volumes

2. Suggest why some reactions that give off gas should only be carried out in a fume cupboard.

---- I don't know urgent help ----

6. A reaction gave off 48 cm³ of carbon dioxide. How many moles of carbon dioxide were produced?

Number of moles = Volume of gas in dm³ / 24
Number of moles = (48 / 1000) / 24
Number of moles = 0.048 / 24
Number of moles = 0.002

7. Mohammed weighed a flask of magnesium and acid as it reacted. He found that it produced 0.00082g of hydrogen gas (H₂).

(a) How many moles of hydrogen gas were produced?

Moles = Mass / Molar Mass
Moles = 0.00082 / (1*2)
Moles = 0.00082 / 2
Moles = 0.00041

(b) How many cm³ of the gas were produced?

Volume in dm³ = moles * 24
Volume in dm³ = 0.00041 * 24
Volume in dm³ = 0.00984
Volume in cm³ = 0.00984 * 1000
Volume in cm³ = 9.84

(edited 8 years ago)

Reply 1

Pigster

If a reaction gives off a gas, it might be a good idea to carry out the experiment in a fume cupboard, especially if the gas is toxic etc, e.g. HCl, NH3, CO, SO2, HCN, H2S, NOx etc.

To be honest, not that many gases that are given off by chemical reactions are especially good for you - even O2 is toxic in large quantities for and extended time.

Reply 2

Bjornhattan

1.2g Sodium / 23.0 g/mol = 0.0522 moles of Sodium

2.8g Salt / 58.5 g/mol = 0.0479 moles of Sodium Chloride

Each mole of NaCl contains one mole of Na and so there are 0.0479 moles of Sodium in Can B. (which is 1.1008g, the same answer as you)

You can see that Can A has more sodium than Can B.

Rounding is the only reason your different methods got different answers, as you may not have used the full answer to all decimal places. If you use the full answer in later calculations it means you don't get errors "rolling over" from calculation to calculation.

Reply 3

Chittesh14

Original post by Bjornhattan

Yeah, sorry, I knew my answers were correct really, just wanted to confirm and find out if people had other ways of doing the same question. However, this is the first time I'm doing C5 lol so that's why I'm confirming everything.

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Reply 4

Chittesh14

Original post by Pigster

Thanks

. But, what is the real reason as to why it should be carried out in a fume cupboard, like why is it a good idea?

Edit: I just found out what a fume cupboard is, so it is carried out there for safety reasons as toxic gases can be emitted into the room and people could suffocate if they are asthmatic and one of the gases is sulfur dioxide or carbon monoxide.

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(edited 8 years ago)

Reply 5

Bjornhattan

Original post by Chittesh14

We've all been there, C5 is tough, frankly most of it is also at AS. But don't worry, everyone finds it hard - and so grade boundaries tend to be quite low for it.

Reply 6

Chittesh14

Original post by Bjornhattan

We've all been there, C5 is tough, frankly most of it is also at AS. But don't worry, everyone finds it hard - and so grade boundaries tend to be quite low for it.

Wow :P. I don't know why, I actually find it quite easy lol. I just love to ask questions to ensure that no matter what question comes up in the test, I will get full marks on every question. That is what I want to aim for - which is why I ask so many questions even if I know they are probably right or easy.

Only topics I find a bit difficult are:

Quantitative Analysis
Titrations - need to do some questions on pH curves
Equilibria - this is by far the hardest topic for me [So be prepared for a thread regarding this soon :biggrin:

]
Ionic equations and precipitation - hopefully this will become easier when I go over the C4 module

I've never ever revised before properly, this is the first time - still I'm barely putting any effort into revision.

Reply 7

Chittesh14

Original post by Bjornhattan

We've all been there, C5 is tough, frankly most of it is also at AS. But don't worry, everyone finds it hard - and so grade boundaries tend to be quite low for it.

Do you have any answers for the questions that don't require calculations such as Q10, Q8/13 and Q15

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(edited 8 years ago)

Reply 8

Bjornhattan

Original post by Chittesh14

Do you have any answers for the questions that don't require calculations such as Q10, Q8/13 and Q15

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Honestly, I don't have any clear cut answers for you. I always used to struggle with the extended answer questions. They seem right though.

Reply 9

Chittesh14

Original post by Bjornhattan

Honestly, I don't have any clear cut answers for you. I always used to struggle with the extended answer questions. They seem right though.

Thanks

. Btw, what year are you in? Would you know about equilibrium?

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Reply 10

Bjornhattan

Original post by Chittesh14

Thanks

. Btw, what year are you in? Would you know about equilibrium?

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Year 12. And yes, I know all about equilibrium, it was actually the topic I understood best in c5.

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Reply 11

Chittesh14

Original post by Bjornhattan

Year 12. And yes, I know all about equilibrium, it was actually the topic I understood best in c5.

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Wow, that's awesome. It's one of the topics I struggle the most on, so I'll be asking questions later tonight if you're alright with it? :smile:

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Reply 12

Bjornhattan

Original post by Chittesh14

Wow, that's awesome. It's one of the topics I struggle the most on, so I'll be asking questions later tonight if you're alright with it? :smile:

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Feel free! I get a bit of a kick out of it to be fair.

Reply 13

Chittesh14

Original post by Bjornhattan

Feel free! I get a bit of a kick out of it to be fair.

Can you mark my answer for Q8 on this paper:

Link to the paper
Mark Scheme Link

My Answer:

As the pressure is increased, the percentage of ammonia yielded increases. For example, at a pressure of 1 atmospheres, the percentage yield was 0.2%, whereas at 50 atmospheres, the percentage yield had increased to 9.5%. This is because increasing the pressure causes a greater likelihood of particle collisions and, thus, shifts the position of equilibrium to the right, in favour of the forward reaction.

On the other hand, as the temperature is increased, the percentage of ammonia yielded decreases. For example, at a temperature of 400 degrees celsius, the percentage yield was 50%, whereas at 550 degrees celsius, the percentage yield decreased to 17%. This is because increasing the temperature moves the position of the equilibrium in the direction of the endothermic reaction. In this case, the forward reaction is exothermic and therefore the backward reaction will be endothermic. In the endothermic reaction, the products are on the left hand side so the equilibrium will move to the left.

Questions:

I don't really understand why the equilibrium moves in the direction of the endothermic reaction when the temperature is increased - why not in the direction of the exothermic reaction?

Also, if I added this to my answer to the question, would it be incorrect or unnecessary?

As the temperature is increased, the percentage yield of ammonia decreases - so less products is made. This means that the equilibriumwill move to the left as the concentration of the reactants will be higher than the concentration of theproducts.

Reply 14

Chittesh14

Original post by Bjornhattan

Feel free! I get a bit of a kick out of it to be fair.

Any ideas buddy? ^

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Reply 15

Bjornhattan

Oh dear, I forgot about this. Sorry.

Original post by Chittesh14

When it comes to Equilibrium, all you have to really know is "le chatelier's principle", which states that balance will always be restored in a closed system. It is much like homeostasis in Biology, if you remember that.

When it comes to pressure, it is not rate of reaction which affects yield, it is actually the gas laws. There are four moles of gas (3+1) on the left, and two moles on the right. So if you increase the pressure, the equilibrium shifts to the right, so you get more ammonia.

You are mostly correct with regard to temperature. What is happening is that when temperature is increased, the reaction tries to "cool itself down", by performing an endothermic process. In this case, that is the backward reaction, so the equilibrium moves.

With regard to your question, if the exothermic reaction was carried out, you'd see the temperature rapidly increase. The exothermic reaction would heat things up, making the situation ever more exothermic.

Reply 16

Chittesh14

Original post by Bjornhattan

Oh dear, I forgot about this. Sorry.

When it comes to Equilibrium, all you have to really know is "le chatelier's principle", which states that balance will always be restored in a closed system. It is much like homeostasis in Biology, if you remember that.

When it comes to pressure, it is not rate of reaction which affects yield, it is actually the gas laws. There are four moles of gas (3+1) on the left, and two moles on the right. So if you increase the pressure, the equilibrium shifts to the right, so you get more ammonia.

You are mostly correct with regard to temperature. What is happening is that when temperature is increased, the reaction tries to "cool itself down", by performing an endothermic process. In this case, that is the backward reaction, so the equilibrium moves.

With regard to your question, if the exothermic reaction was carried out, you'd see the temperature rapidly increase. The exothermic reaction would heat things up, making the situation ever more exothermic.

Oh my God, thank you so much. I actually get it now and how it relates to the Le Chatelier's concept.
When the temperature is increased, as Le Chatelier's says - the reaction tries to do the opposite, decrease the temperature and cool itself down. So, obviously, it will perform an endothermic process as it will release heat. The endothermic reaction is backwards and therefore the equilibrium moves to the left.

What if I say this, would it be right?

As the temperature is increased, the percentage yield of ammonia decreases - so less products is made. This means that the equilibriumwill move to the left as the concentration of the reactants will be higher than the concentration of theproducts.

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(edited 8 years ago)

Reply 17

Bjornhattan

Original post by Chittesh14

No, the equilibrium would in that case move to the right, because more products would be made to counteract the decrease.

However, you do not need to go to that level of detail.

Reply 18

Chittesh14

Original post by Bjornhattan

No, the equilibrium would in that case move to the right, because more products would be made to counteract the decrease.

However, you do not need to go to that level of detail.

Oh right yeah, thanks :smile:

.

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Reply 19

Chittesh14

Original post by Bjornhattan

No, the equilibrium would in that case move to the right, because more products would be made to counteract the decrease.

However, you do not need to go to that level of detail.

Hey, I don't really understand this part, can you please explain what it means.

"When a balance is reached between the amount of reactions and products present, the reaction has reached equilibrium."

Does this mean that the reaction will reach equilibrium when the rate of the forward and backward reactions is equal.
Also, does it mean that the products on both sides of the equations - forward and backward will be equal too or does 'balanced' mean something else?
If it means they are equal, does it mean the reactants and products for both equations will be the same?

For example, the forward reaction has 30% reactants and 70% products and the backward reaction now has 30% reactants and 70% producers if they are at the same rate.

Also, what if I say this instead of what I said previously,

As the temperature is increased, the percentage yield of ammonia decreases - so less products is made. This means that the equilibrium will be on the left as the concentration of the reactants will be higher than the concentration of the products. Now, as Le Chatelier's states, the equilibrium will shift in a way to reduce the change by moving to the right to increase the yield of ammonia.

Also, one question:

When the pressure is increased, the equilibrium will try to reduce the effect of the change, so it will move to the right to reduce the pressure as there are fewer particles on the right.
So, basically, the equilibrium will move in the direction where there are fewer particles or lower pressure.
So, if there were less moles on the left hand side of the equation, the equilibrium would move left to reduce the pressure as there is lower pressure on the left of the equation?

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(edited 8 years ago)