Chemistry: Post your doubts here!

[Metanoia]

View attachment 51086

Form equation first

2NaCl + 2H2O --> 2NaOH + Cl2 + H2

Then its a simple mole calculation question involving 58.5 kg of NaCl.

 

[rk_butterfly]

chemistry 9701 topical questions plzzzz
urgent

 

[My Name]

chemistry 9701 topical questions plzzzz
urgent

Check here
https://www.xtremepapers.com/community/threads/as-chemistry-topical-finally-here.23520/
(PS only MCQ's answers are there :/)

 

[princess Anu]

Temperature increase (due to heat of neutralization) should still be 6 degree celsius.

Expt 1: 50cm3 of acid + 50 cm3 of alkali (0.05 mol of water formed, releasing heat)
Expt 2: 100cm3 of acid + 100 cm3 of alkali (0.1 mol of water formed, releasing heat)

Basically you are generating 2 twice the amount of heat in expt 2, but its used to heat up twice the volume of solution. So temperature rise is the same in both experiments.

More importantly in this question, mixing any equal volumes of acid and alkali would give you a 6 degree celsius rise.

The reason being? if we use for example 50/4 of acid and 50/4 for alkali temperature rise will be 3 ,no?

 

[princess Anu]

Form equation first

2NaCl + 2H2O --> 2NaOH + Cl2 + H2

Then its a simple mole calculation question involving 58.5 kg of NaCl.

I don't understand why making this equation? shouldn't we make the anode cathode equations? :S

 

[Metanoia]

The reason being? if we use for example 50/4 of acid and 50/4 for alkali temperature rise will be 3 ,no?

Generally ,

heat released during neutralization = heat used to raise temperature of solution

moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change

From expt 1,
(0.05) x heat of neutralization = 100 x heat capacity of solution x temperature change of expt 1

From expt 2,
(0.1) x heat of neutralization = 200 x heat capacity of solution x temperature change of expt 2

Take the eqn 2 and divide it by eqn 1

(0.1) x heat of neutralization ..........200 x heat capacity of solution x temperature change of expt 2
--------------------------------- =.... -----------------------------------------------------------------------
(0.05) x heat of neutralization .......100 x heat capacity of solution x temperature change of expt 1

You will end up with,
temp change of expt 1 = temp change of expt 2

 

[Metanoia]

I don't understand why making this equation? shouldn't we make the anode cathode equations? :S

Hmm.. why wouldn't you want to make use of the supplied info that reactants are brine (NaOH + H2O) and products are NaOH, H2 and Cl2?

If you were to create your own cathode /anode equations, what did you end up with?

 

[princess Anu]

Generally ,

heat released during neutralization = heat used to raise temperature of solution

moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change

From expt 1,
(0.05) x heat of neutralization = 100 x heat capacity of solution x temperature change of expt 1

From expt 2,
(0.1) x heat of neutralization = 200 x heat capacity of solution x temperature change of expt 2

Take the eqn 2 and divide it by eqn 1

(0.1) x heat of neutralization ..........200 x heat capacity of solution x temperature change of expt 2
--------------------------------- =.... -----------------------------------------------------------------------
(0.05) x heat of neutralization .......100 x heat capacity of solution x temperature change of expt 1

You will end up with,
temp change of expt 1 = temp change of expt 2

Thanks alot
one more question,^ this thing applies for this Example because molar ratio of reactant to product is 1:1? or does it always apply when the same reaction is repeated but with different volumes of reactants? + the formula i.e heat capacity of solution x temperature change * mass gives heat change for 1 mole?

 

[princess Anu]

n

Hmm.. why wouldn't you want to make use of the supplied info that reactants are brine (NaOH + H2O) and products are NaOH, H2 and Cl2?

If you were to create your own cathode /anode equations, what did you end up with?

Umm I couldn't do it

 

[princess Anu]

for this question part b ii Marking scheme doesn't mention Kmno4, I don't understand why? Alcohols do react with kmn04 to give carboxylic acid right?

 

[danial 234]

for this question part b ii Marking scheme doesn't mention Kmno4, I don't understand why? Alcohols do react with kmn04 to give carboxylic acid right?

Actually, acidified sodium/potassium dichromate is used for oxidation of alcohols. Because, KMnO4 especially acidified KMnO4 is a very powerful oxidising agent, and it can even break carboon-carbon bonds. So, dichromate solution is used for oxidation of alcohols. Though CIE uses KMnO4 in a number of mark schemes and question papers, but it is better to be on a safe side , by writing K2Cr2O7.

 

[yoyo00xx]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_13.pdf

please help me with q31 The Chill Master

 

[Metanoia]

Thanks alot
one more question,^ this thing applies for this Example because molar ratio of reactant to product is 1:1? or does it always apply when the same reaction is repeated but with different volumes of reactants? + the formula i.e heat capacity of solution x temperature change * mass gives heat change for 1 mole?

The equation: moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change

Can be used with different volumes, concentrations and reactants, it is not necessary that the molar ratio of reactant to water is 1:1.

For different concentrations and volumes of acids + alkali , what is important is to use mole concepts to find out the moles of limiting reactant (which could either be the acid or alkali), this will determine the moles of water that would be formed.

For example
Mixing 40cm3 of 0.1 mol/dm3 of H2SO4 with 100cm3 of 0.1 mol/dm3 NaOH

H2SO4 + 2NaOH --> Na2SO4 + 2H2O
1) work out that there is 0.04 mol of H2SO4 and 0.1 mol of NaoH
2) work out that the H2SO4 is the limiting reactant (0.04 mol)
3) work that therefore, 0.08 mol of water would be formed.

moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change
0.08 x heat of neutralization = 140 x heat capacity of solution x temperature change

Once you have the mole of water formed, then you can use the equation

n

Umm I couldn't do it

For this question, you do not have to do it. I will show you how it could be done for discussion sake.

Cathode:
2H2O + 2e- -> H2 + 2OH-

Anode
2Cl- --> Cl2 + 2e-

Merging cathode and anode reaction:
2Cl- + 2H2O --> H2 + Cl2 + 2OH-

Tidying up to include sodium on both sides:
2Na+ + 2Cl- + 2H2O --> H2 + Cl2 + 2Na+ + 2OH-

Which gives us the balance equation hinted by the question:
Brine --> hydrogen + chlorine + sodium hydroxide
2NaCl + 2H2O --> Cl2 + H2 + 2NaOH

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_13.pdf

please help me with q31 The Chill Master

If you are having difficulty figuring out the oxidation states of N and Cl inside NH4ClO4, view them in this manner

NH4+ and ClO4-

For N in NH4+ :
N + 4(+1) = + 1
N = -3

For Cl in ClO4- :
Cl + 4(-2) = -1
Cl = +7

 

[princess Anu]

The equation: moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change

Can be used with different volumes, concentrations and reactants, it is not necessary that the molar ratio of reactant to water is 1:1.

For different concentrations and volumes of acids + alkali , what is important is to use mole concepts to find out the moles of limiting reactant (which could either be the acid or alkali), this will determine the moles of water that would be formed.

For example
Mixing 40cm3 of 0.1 mol/dm3 of H2SO4 with 100cm3 of 0.1 mol/dm3 NaOH

H2SO4 + 2NaOH --> Na2SO4 + 2H2O
1) work out that there is 0.04 mol of H2SO4 and 0.1 mol of NaoH
2) work out that the H2SO4 is the limiting reactant (0.04 mol)
3) work that therefore, 0.08 mol of water would be formed.

moles of water formed x heat of neutralization = mass of solution x heat capacity of solution x temperature change
0.08 x heat of neutralization = 140 x heat capacity of solution x temperature change

Once you have the mole of water formed, then you can use the equation


For this question, you do not have to do it. I will show you how it could be done for discussion sake.

Cathode:
2H2O + 2e- -> H2 + 2OH-

Anode
2Cl- --> Cl2 + 2e-

Merging cathode and anode reaction:
2Cl- + 2H2O --> H2 + Cl2 + 2OH-

Tidying up to include sodium on both sides:
2Na+ + 2Cl- + 2H2O --> H2 + Cl2 + 2Na+ + 2OH-

Which gives us the balance equation hinted by the question:
Brine --> hydrogen + chlorine + sodium hydroxide
2NaCl + 2H2O --> Cl2 + H2 + 2NaOH

thanks
I'm assuming the equation i.e heat of neutralization = mass * .. is for 1 mole of water only right

 

[My Name]

 

[My Name]

Answers
B
C
B

 

[The Chill Master]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_13.pdf

please help me with q31 The Chill Master

I am confused b/w A & B cuz 1st 2 r correct 3rd one i can't determine.Because oxidation in terms of removal of hydrogen atom is taking place :/

 

[Metanoia]

thanks
I'm assuming the equation i.e heat of neutralization = mass * .. is for 1 mole of water only right

Hmm...yes. because simply,
1 x heat of neutralization = mass x specific heat capacity x temp rise

 

[Metanoia]

View attachment 51114

One H2 adds across each C=C bonds, same reasoning for Br2.

So 2 moles for both.

View attachment 51115
Answers
B
C
B

Which years are these from?