Chemistry: Post your doubts here!

[Metanoia]

Hi,
I do not understand why there will be 42 hydrogen atoms in the first place. The rest of it makes a lot of sense. Thank you.

for a saturated molecule: CnH2n+2

 

[kruti]

Ahhh alright. Now it seems so easy! Thanks Metanoia

 

[Metanoia]

Q6 guys

hmmm, it was actually answered a page ago.

s08qp1

Q6.
H2O(s) --> H2O(g)

Moles of H2O = mass of ice/Mr of H2O = 1/18= 0.0556

PV/nT (ideal gas) = PV/nT (H2O)
(1)(24)/(1)(298) = (1)V/(0.0556)(596)

V = 2.67 dm^3

 

[kruti]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf

question 2 ans. A
I am getting the answer but my school teacher says my method is wrong. I found the empirical formula using percentage composition by mass. what's the other way?

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
Number 8 ?? Answer D . If u can do this and explain it ur a genius! Hardest question! Please anyone explain it

Ca(s) --> Ca(g)--> Ca+ (g) --> Ca2+ (g) --> Ca2+ (aq)

add up the energy changes involved in the stages, with 1st and 2nd IE from the data booklet.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf

question 2 ans. A
I am getting the answer but my school teacher says my method is wrong. I found the empirical formula using percentage composition by mass. what's the other way?

s08qp1

Just use proportion if its easier to visualise.

mass of N in 100 g of fertiliser = 15 g

mass of N in 14 g of fertiliser = (15/100) x 14= 2.1 g

moles of N in 14 g of fertiliser = 2.1 /14 = 0.15 mol

conc of N in 5 dm^3 of solution = 0.15/5 = 0.03 mol/dm^3

 

[kruti]

s08qp1

Just use proportion if its easier to visualise.

mass of N in 100 g of fertiliser = 15 g

mass of N in 14 g of fertiliser = (15/100) x 14= 2.1 g

moles of N in 14 g of fertiliser = 2.1 /14 = 0.15 mol

conc of N in 5 dm^3 of solution = 0.15/5 = 0.03 mol/dm^3

Thanks. Also, i want to know whether my method is wrong coz I am getting the right answer too.

 

[Metanoia]

Thanks. Also, i want to know whether my method is wrong coz I am getting the right answer too.

Are you able to post your working?

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 answer is B
Q14 answer A what about C??
Q19 answer is B i don't get how III is included?
Q26 shouldn't the answer be A like i was so sure isn't X a nitrile and we do hydrolysis with dilute to get carboxylic acid? answer is C
Q28 i know its an easy one but when we add H2 in the presence of Ni catalyst we dont change an aldehdy to alcohol do we? answer is A
Q32 how do we deduce 2 using that info?? answer is A
Q33 how 3??? answer is C
Q37 i have no idea answer B
Please help

 

[Ahmed Aqdam]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 answer is B
Q14 answer A what about C??
Q19 answer is B i don't get how III is included?
Q26 shouldn't the answer be A like i was so sure isn't X a nitrile and we do hydrolysis with dilute to get carboxylic acid? answer is C
Q28 i know its an easy one but when we add H2 in the presence of Ni catalyst we dont change an aldehdy to alcohol do we? answer is A
Q32 how do we deduce 2 using that info?? answer is A
Q33 how 3??? answer is C
Q37 i have no idea answer B
Please help

14: Ionisation energy decreases down the group.
19: The molecule is CH3-CHI-CH2I
26: Halogenalkane react with NaCN not HCN
28: H2 in presence of Ni catalyst also reduces aldehyde to alcohol.
32: As CO burns readily to form CO2 so CO2 is more stable in which carbon is+4.
Formation of CO2 is formation of CO and also combustion of CO so combustion of CO is negative so formation of CO2 is more negative.
As more CO2 is more produced so equilibrium constant is high.
33: More moles of gas are produced so one mole occupies 24dm^3 so more volume.
37: Replace D with H and then look at the reactions.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Q9 answer is B
Q14 answer A what about C??
Q19 answer is B i don't get how III is included?
Q26 shouldn't the answer be A like i was so sure isn't X a nitrile and we do hydrolysis with dilute to get carboxylic acid? answer is C
Q28 i know its an easy one but when we add H2 in the presence of Ni catalyst we dont change an aldehdy to alcohol do we? answer is A
Q32 how do we deduce 2 using that info?? answer is A
Q33 how 3??? answer is C
Q37 i have no idea answer B
Please help

w07qp1

Qn 9 :
moles of SO3 2- : moles of electrons : moles of metal
0.0025 : ? : 0. 005
1 :? : 2
1 : 2 : 2

2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q14. IE should be decreasing down the Group.
Q26. We use HCN (with trace NaCN) to step up carbonyls. We use NaCN or KCN to step up halogen alkanes.

Q19. Put the iodine atoms on 1st and 2nd carbon. CH2ICHICH3, chiral carbon is the middle carbon.

Q28. Ketones and aldehydes can be reduced by H2 to alcohols. Only COOH wouldn't be reduced by H2.

Q32. Burns readily means CO --> CO2 exothermic.
1. CO2 more stable than CO in terms of energy
2. as explained, exothermic reaction
3. ratio of CO2 is high compared to CO, therefore Kc is high

Q33.
Statement 1: (False) The gas is exerting the same pressure as the atmosphere outside, otherwise the plunger will either move left or right.
Statement 2: (True) When plunger moves inwards, it causes the pressure to increase, the eqm will shift to the left to reduce pressure (produce less moles of gas).
Statement 3: (True) For every one mole of PCl5 dissociated, two moles of gases were formed (1 mol of PCl3 and 1 mol of Cl2). So total volume of gases after dissociation is more than original volume of gas without dissociation.

Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Help needed in question 6,15,
22 how do YOU KNOW??!!
27

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Help needed in question 6,15,
22 how do YOU KNOW??!!
27

26 as well please!!!
Thank you

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Help needed in question 6,15,
22 how do YOU KNOW??!!
27

s08qp1

Q6. Answered in post 9750 just above.

H2O(s) --> H2O(g)

Moles of H2O = mass of ice/Mr of H2O = 1/18= 0.0556

PV/nT (ideal gas) = PV/nT (H2O)
(1)(24)/(1)(298) = (1)V/(0.0556)(596)

Q15. CaCo3 --> CaO + CO2
mass of CaCO3 = 1200 million tonnes
work out moles of CaCO3
then moles of CO2
then mass of CO2

Q22. The carbons are all saturated with 4 single bonds, so tetrahedral, 109.5

Q27.
Use the 4 options and trial and error.

Before dehydration, the C=C bonds were actually CH-COH.

Check which of the 4 could possibly give us a tertiary alcohol (resistant to oxidation) before dehydration.
View attachment 44341
In this case, it was D.

 

[♣♠ Magnanimous ♣♠]

explain me ques 5 and 6

 

[ZaqZainab]

w07qp1

Qn 9 :
moles of SO3 2- : moles of electrons : moles of metal
0.0025 : ? : 0. 005
1 :? : 2
1 : 2 : 2

Q14. IE should be decreasing down the Group.
2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

Q26. We use HCN (with trace NaCN) to step up carbonyls. We use NaCN or KCN to step up halogen alkanes.

Q19. Put the iodine atoms on 1st and 2nd carbon. CH2ICHICH3, chiral carbon is the middle carbon.

Q28. Ketones and aldehydes can be reduced by H2 to alcohols. Only COOH wouldn't be reduced by H2.

Q32. Burns readily means CO --> CO2 exothermic.
1. CO2 more stable than CO in terms of energy
2. as explained, exothermic reaction
3. ratio of CO2 is high compared to CO, therefore Kc is high

Q33.
Statement 1: (False) The gas is exerting the same pressure as the atmosphere outside, otherwise the plunger will either move left or right.
Statement 2: (True) When plunger moves inwards, it causes the pressure to increase, the eqm will shift to the left to reduce pressure (produce less moles of gas).
Statement 3: (True) For every one mole of PCl5 dissociated, two moles of gases were formed (1 mol of PCl3 and 1 mol of Cl2). So total volume of gases after dissociation is more than original volume of gas without dissociation.

Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2

Thanks
i am still confused about
Q9 how did you find the moles? shouldn't metallic salt be 0.10/(50*10^-3)=2??
SO3 2- be 0.10/(25*10^-3)=4??

14: Ionisation energy decreases down the group.
19: The molecule is CH3-CHI-CH2I
26: Halogenalkane react with NaCN not HCN
28: H2 in presence of Ni catalyst also reduces aldehyde to alcohol.
32: As CO burns readily to form CO2 so CO2 is more stable in which carbon is+4.
Formation of CO2 is formation of CO and also combustion of CO so combustion of CO is negative so formation of CO2 is more negative.
As more CO2 is more produced so equilibrium constant is high.
33: More moles of gas are produced so one mole occupies 24dm^3 so more volume.
37: Replace D with H and then look at the reactions.

Thanks
Q19 why not
CH3-CH2-CHI2??

 

[Metanoia]

Thanks
i am still confused about
Q9 how did you find the moles? shouldn't metallic salt be 0.10/(50*10^-3)=2??
SO3 2- be 0.10/(25*10^-3)=4??


Thanks
Q19 why not
CH3-CH2-CHI2??

Moles = conc x volume

you did it as moles = conc/volume ?

 

[Metanoia]

View attachment 44961 explain me ques 5 and 6

Q5. Fix P4O10 as 1 mole

P4O10 + 6CaO + --> 2Ca3(PO4)2

Q6.
Just use proportion if its easier to visualise.

mass of N in 100 g of fertiliser = 15 g

mass of N in 14 g of fertiliser = (15/100) x 14= 2.1 g

moles of N in 14 g of fertiliser = 2.1 /14 = 0.15 mol

conc of N in 5 dm^3 of solution = 0.15/5 = 0.03 mol/dm^3

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q4 D is the answer
Q8 D is the answer what about A? PV=nRT n and R and T constant so P=1/V so P increases V decreases? and i do understand why D is correct but why is A wrong?
Q22 i have a problem with angles any notes will do?
Q26 C or B isnt it the same thing C is the actual answer
Q29 what about C? D is the answer
Q38 why int 2 correct
Q39 it is a primary alcohol so shouldn't it be forming an aldehyde

 

[ZaqZainab]

Moles = conc x volume

you did it as moles = conc/volume ?

my bad Thanks again