A few past paper questions (1 Viewer)

[Danni07]

My chemistry half yearly is on Wednesday and in preparation I've been doing some past papers. However, I've encountered a few problems with the 2001 paper. I was wondering if someone would mind having a look at my answers and working out, and let me know what I'm doing wrong?

Question 17:
Students were asked to perform a first hand investigation to determine the molar heat of combustion of ethanol.
Lab Data:
Mass of water = 250.0g
Initial mass of burner = 221.4g
Final mass of burner = 219.1g
Initial temperature of water = 19.0 deg
Final temperature of water = 59.0 deg

c) Calculate the molar heat of combustion using the student's data.


∆H = mC∆T
∆H = 0.0023kg x 4.18x103 j/kg x 40°c
∆H = 384.56j/kg

Molar mass of ethanol = 34g (0.034 kg)

Molar heat = 384.56/0.034
= 11310.6j/mol

Obviously that isn't right, I know my problem has something to do with the units, I just don't know what?

also, Question 18
A galvanic cell was made by connecting two half cells. One half cell was made by putting a copper electrode in copper (II) nitrate solution. The other half cell was made by putting a silver electrode in a silver nitrate soulution. The electrodes were connected by a voltmeter.

b) Using the standard potentials table in the data sheet. calculate the theoretical voltage of the galvanic cell

ANODE - Cu(s) → Cu2+ + 2e- (-0.34V)
CATHODE - Ag+ + 2e- → Ag(s) (0.80V)

Total cell = Potential of cathode - Potential of anode
Total voltage = 1.14V

This answer could be right, I'm not sure and I'd really appreciate if someone could check it for me.
Thanks heaps in advance.

 

[table for 1]

Question 17:
Students were asked to perform a first hand investigation to determine the molar heat of combustion of ethanol.
Lab Data:
Mass of water = 250.0g
Initial mass of burner = 221.4g
Final mass of burner = 219.1g
Initial temperature of water = 19.0 deg
Final temperature of water = 59.0 deg

c) Calculate the molar heat of combustion using the student's data.


∆H = mC∆T
∆H = 0.25kg x 4.18x103 j/kg x 40°c
∆H = 384.56j/kg

[EDIT: since 'm' refers to the mass of the water]

the formula ∆H = -mC∆T has the negative in front, to indicate whether the reaction was endothermic or exothermic. i don't know if you need to write the negative if you're not finding the type of reaction though. i would write it, just to be on the safe side. since chem markers are so pedantic - they take a mark off for no states on the equations

 

[Danni07]

ohh... heh i get it now *blushes* thank you

EDIT: *pulls hair out* is the final answer 1229.41kj/mol? Becuase I keep coming up with strange answers.

 

[table for 1]

also, i didn't notice this before but the heat produced in KJ/mol is found by the formula
heat produced = {∆H in KJ}/{moles of fuel used}
you can't do what you did because you divided the ∆H with molar mass. you must convert the mass of ethanol used into moles first. ie. n=mass/molar mass

 

[Danni07]

ah ok, that's why they give you that information. Thanks so much for all your help.

 

[table for 1]

you're welcome

good luck in your test. mine is coming up too.

oh, i still have to check Q18. i suck at galvanic cells, but i think your answer is right. the only thing you need to change is the cathode equation. it should be:
2Ag+ + 2e- → 2Ag(s) (0.80V)
because you hadn't balanced it yet. [but do not change the E°]

 

[Casmira]

Oh shit you reminded me in the exam, I had silver nitrate and zinc nitrate ... I only did 1 electron on silver side

*bashes head on desk*

 

[Danni07]

I've come across another difficult question, this time in the 2002 paper, and I really dont know how to answer it.

Question 22
solutions of hydrochloric acid, acetic acid sulfuric acid were prepared. Each of the solutions had the same concentration (0.01 mol/L). The pH of the acetic acid was 3.4.

a) Calculate the pH of the hydrochloric acid solution.
The only way i know to work out the pH of a solution is to know the hydronium ion concentration, which this question doesn't give me. Any ideas?

b)Compare the pH of the sulfuric acid to the pH of the hydrochloric acid solution. Justify your answer. (No calculations are necessary)
Maybe once i can work out the answer to a, this will make more sense...? How can I compared it though without knowing what the pH of the sulfuric acid is?

Any help would be greatly appreciated, once again.

 

[Skywalker]

a) They give you the hydronium ion concentration - it's there in brackets (0.01mol/L). Assuming the HCL is completely ionised, then it's [H+] conc. is equal to 0.01mol/L. Therefore, pH = 10^-0.01

b) Sulphuric acid is diprotic, HCL is monoprotic - you need multiply the [H+] concentration of sulphuric acid by two. Even without using a calculator, you'll know that the pH of sulphuric acid is lower than HCL

 

[funking_you]

Question 22
solutions of hydrochloric acid, acetic acid sulfuric acid were prepared. Each of the solutions had the same concentration (0.01 mol/L). The pH of the acetic acid was 3.4

Important Notes:
- we have three acids; 2 of which are 'strong', i.e. hydrochloric and sulfuric, and one which is weak; acetic.
- a 'strong' acid is defined as one that completely ionisies in solution, i.e
HCl(aq) --> H⁺(aq) + Cl^-(aq)
H₂SO₄(aq) --> 2H⁺(aq) + SO₄^2-(aq)
- all the acids are 0.01 mol/L, hence for a HCl, being a strong acid, the H⁺ will also be 0.01 mol (since there is a 1:1 ratio between HCl and H⁺)
- for sulfuric acid, the H⁺ will be 0.02 mol/L...why? well as shown by the reaction above, since it is a strong acid, it completely ionises into H⁺ ions, and after we balance it, we notice there is a 1:2 molar ratio between the acid and the hydrogen ions.
- since acetic acid is weak, we dont know how much it ionises, but we do know it ionises only a little bit - but since we are give its pH (3.4) we can apply the formula pH=-log(H⁺) to find H⁺ concentration, which is 4 x 10^-4 mol/L or 0.0004 mol/L.

- notice that the H⁺ concentration of sulfuric acid is double that of hydrochloric acid, well you can see why in the chemcial reactions given, we say that sulfuric acid is a 'diprotic' acid, since it will ionise to form 2 moles of H⁺ ions for every one mole of acid.


That answers to both the questions lie in all that imformation.

Best of Luck,
George

 

[Danni07]

Thank you both for your answers, they were very helpful.