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Chemistry: Post your doubts here!

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PLZ someone help with part (II) (III) and (IV) PLZZ !!!
i may not be right but let me try
see if u add Na2SO4 on the left hand electrode, its going to dissociate this way
Na2SO4 _____ 2Na*+ + SO4*2-
the solution in the beaker are this way
fe2so4 = fe2+ +so4*2-
feso4 = fe*2 + so4*2-
an increase in conc of so42- will shift the equlibrium of both the ions towards the left so the conc. of fe2+ and fe3+ increases in the same way therefor no change in emf

however in the case of agso4, the solution is as follows
agso4= ag+ +so42-
increase in sulphate conc will shift the equilibrium towards the left decreaing the concentration of ag+ so the Eelectrode will decease decreasing the overall emf
 
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Hi friends pls can anyone help me to understand how the Ksp is calculated here.....Oct 10 p41
 

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Hello, can you please tell me if it's possible to finish paper 5 in a week? given that at the beginning of said week I never opened any paper and at the end of the same week is the actual exam i.e. 30 October - 5 November?
 
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Thanks dear for your answer, but I was specifically needing help for no. 2, I don't know how to make [Pb2+] equal to [Cl-]. I hope you understand when the power is 1 for the concentration I can calculate it easily but I am having problem when there is a power greater than 1...........waiting for your response
 
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Thanks dear for your answer, but I was specifically needing help for no. 2, I don't know how to make [Pb2+] equal to [Cl-]. I hope you understand when the power is 1 for the concentration I can calculate it easily but I am having problem when there is a power greater than 1...........waiting for your response

Let concentration of eqm concentration of Pb2+ be x

[Cl-] = 2x

Ksp = [Pb2+][Cl-]^2
= (x) (2x)^2
= (x)(4x^2)
= 4x^3
 
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Hi can pls help me with this problem. 9701/41/O/N/2012 qstn 1 b I. Why cannot we choose any equation for MnO4- to solve the problem....I chosed the one with E০ 1.67V, but the MS worked this problem out with the one having a value of 1.52V....... I have attached the pics........IMG_20151005_095854_635.jpg IMG_20151005_095933_937.JPG
 
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Hi can pls help me with this problem. 9701/41/O/N/2012 qstn 1 b I. Why cannot we choose any equation for MnO4- to solve the problem....I chosed the one with E০ 1.67V, but the MS worked this problem out with the one having a value of 1.52V....... I have attached the pics........View attachment 57150 View attachment 57151

Short answer:

1.67V is for neutral/mildly alkaline conditions
1.52V is for acidic conditions.

In case you are wondering, the 0.56V half equation is not commonly needed as the MnO42- is unstable and tends to undergo disproportionation.
 
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(b) A mixture containing 0.50 mol of CO2, 0.50 mol of H2, 0.20 mol of CO and 0.20 mol of H2O was placed in a 1.0 dm3 flask and allowed to come to equilibrium at 1200 K. Calculate the amount, in moles, of each substance present in the equilibrium
mixture at 1200 K.

Can someone please help me with this question?
 
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chemistry MCQ may/june 2015 varient 2 question 2

The shell of a chicken’s egg makes up 5% of the mass of an average egg. An average egg has a mass of 50 g.
Assume the egg shell is pure calcium carbonate.
How many complete chicken’s egg shells would be needed to neutralise 50 cm3 of 2.0 mol dm–3 ethanoic acid?
A 1 B 2 C 3 D 4

I assumed its D buts its B and how?
 
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(b) A mixture containing 0.50 mol of CO2, 0.50 mol of H2, 0.20 mol of CO and 0.20 mol of H2O was placed in a 1.0 dm3 flask and allowed to come to equilibrium at 1200 K. Calculate the amount, in moles, of each substance present in the equilibrium
mixture at 1200 K.

Can someone please help me with this question?
For this we need the value of Kc at 1200K temperature. It will be mentioned somewhere near the equation. Use that to solve this problem, if u dont know how to solve it, just temme the Kc value, i will explain you... Also the balanced equation.
 
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chemistry MCQ may/june 2015 varient 2 question 2

The shell of a chicken’s egg makes up 5% of the mass of an average egg. An average egg has a mass of 50 g.
Assume the egg shell is pure calcium carbonate.
How many complete chicken’s egg shells would be needed to neutralise 50 cm3 of 2.0 mol dm–3 ethanoic acid?
A 1 B 2 C 3 D 4

I assumed its D buts its B and how?
This is what I did in my exams,
CaCO3 + 2CH3COOH ----> (CH3COO)2Ca + CO2 + H2O
The shell of a chicken’s egg makes up 5% of the mass of an average egg and average mass is 50g
moles of egg shell = 5% of 50/100 = 0.025mol
moles of ethanoic acid = 50*10^-3 * 2 = 0.1. According to stoichiometry, 0.1/2 = 0.05mol
We are asked How many complete chicken’s egg shells would be needed to neutralise 50 cm3 of 2.0 mol dm–3 ethanoic acid?
So it will be moles of CaCO3 / moles of CH3COOH = 2
 
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For this we need the value of Kc at 1200K temperature. It will be mentioned somewhere near the equation. Use that to solve this problem, if u dont know how to solve it, just temme the Kc value, i will explain you... Also the balanced equation.

Value of Kc is 1.44
The reaction is CO2 + H2 ----> CO + H2O
 
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q for x chem.png
I know how to make compounds D,E and F. But what is the point of knowing that when I don't even know how to make A, B and especially C.
Yeah, so I need to know how to make compounds A, B and C.
 
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