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Chemistry: Post your doubts here!

FarahMJ

FarahMJ

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Can you please help me with these as well
0451d-697b8e7b-fced-465c-8719-e469983a6044.png

0451d-8c466ae4-f25c-4c50-b994-f7818cda2cc0.png


Answers are B and D
 
MariamMalik

MariamMalik

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Metanoia said:
Q1. Find out the number of moles of O2, since 1 mole of gas occupies 24000 cm^3 at RTP.

moles of O2 = 500/24 000
molecules of O2 = moles x 6.02 x 10^23 = (500/24000) x 6.02 x 10^23 = 1.25 x 10^22

Q2. Comparing the original structure to the resulting structure, we see that 5 C=C double bonds "gone".
So 5 moles of H2 are added.

If unable to visualise the above explanation, the diagram can show how H could be added in these locations.
View attachment 44133

Q7 and Q10

View attachment 44134

Q21. Is the answer really D? I got B.
View attachment 44136

Q22. The C=C undergoes oxidative cleaving, and the ends are oxidised to COOH.

View attachment 44137

Q23.
If confident, this question can be approached in a mathematical way

CnH(2n+2) + (3n+1)/2 O2 --> n CO2 + (n+1)H2O

when n increases, moles of O2 increases linearly. So its a line with a positive gradient.
Click to expand...

Yeah the answer in the mark scheme is D, I think it should be B too. And thank you so much!
 
Metanoia

Metanoia

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FarahMJ said:
Can you please help me with these as well
0451d-697b8e7b-fced-465c-8719-e469983a6044.png

0451d-8c466ae4-f25c-4c50-b994-f7818cda2cc0.png


Answers are B and D
Click to expand...

Q15. This is a similar to question Q14 of 2006.

Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)
1.32 g is mass of NO2 and O2 gas, the left over mass 0.68 g is XO.

Mass of XO = 0.68g
Moles of XO = 0.68/ (Mr of X + 16)

Mass of X(NO3)2 = 2 g
Moles of X(NO3)2 = 2/(Mr of X + 124)

since moles of X(NO3)2 = moles of XO
2/(Mr of X + 124) = 0.68/ (Mr of X + 16)

We can solve for Mr of X using the normal maths approach, but it might be faster to do trial and error from the Mr of the four options.
Be : 2/(9 + 124) does not equal to 0.68/ (9 + 16)
Ca: 2/(40 + 124) = 0.68/ (40+ 16)
Therefore, answer is Ca.

Q17. At RTP, 1 mol of gas (assume ideal) occupies 24 000 cm3.

mole of O2 = 300 /24 000 = 0.0125 mol

2Ca + O2 --> 2 CaO
2Mg + O2 --> 2 MgO
4K + O2 --> 2K2O
4Na + O2 --> 2 Na2O

0.0125 mol of O2 will react with:
0.025 mol of Ca (1 g)
0.05 mol of Mg (1.2 g)
0.05 mol of K (1.95 g)
0.05 mol of Na (1.15 g)
 
Last edited:
Metanoia

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ZaqZainab said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_1.pdf
Q9
Q35 isn't 3 suppose to be correct too?
Q37
Click to expand...

Q9. HCl is a strong acid, so 1M of HCl is expected to produce 1 M of H+ ions

A. CH3COOH is a weak acid, so 1 M of CH3COOH produces less than 1 M of H+ ions
B. HNO3 is a strong acid, so produces 1 M of H+ ions
C. NaOH is a strong base, so produces 1 M of OH- ions
D. H2SO4 is dibasic, 1M of H2SO4 produces more than 1M of H+, but slightly less than 2M of H+


Q35. Carbon monoxide is neutral, and does not react like the other two acidic gases.

Q37.
Statement 1: Correct. The carbon atoms have 3 bond pairs and 0 lone pairs, so 120 degree angle along the molecule.
Statement 2: Correct. Electrophilic addition at the C=C area
Statement 3: Correct. Nucleophilic addition at the C=O area
 
Metanoia

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Last edited:
immie.rose

immie.rose

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Abdel Moniem said:
number of moles of Tl+NO3-=10*0.30/1000
=3*10^-3
Number of moles of NH4+VO3-=20*0.10/1000
=2*10-3
Divide by the smalles figure, we would obtain the ratio of 1.5:1 *2 to get rid of a fraction so the mole ration will 3:2 so for every 3 moles of Tl+( six electrons are removed) only two moles are reduced so divide the six electrons by 2 to figure out how many electrons are gained per mole
Click to expand...
Thanks!!
 
FarahMJ

FarahMJ

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Metanoia said:
Q15. This is a similar to question Q14 of 2006.

Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)
1.32 g is mass of NO2 and O2 gas, the left over mass 0.68 g is XO.

Mass of XO = 0.68g
Moles of XO = 0.68/ (Mr of X + 16)

Mass of X(NO3)2 = 2 g
Moles of X(NO3)2 = 2/(Mr of X + 62)

since moles of X(NO3)2 = moles of XO
2/(Mr of X + 124) = 0.68/ (Mr of X + 16)

We can solve for Mr of X using the normal maths approach, but it might be faster to do trial and error from the Mr of the four options.
Be : 2/(9 + 124) does not equal to 0.68/ (9 + 16)
Ca: 2/(40 + 124) = 0.68/ (40+ 16)
Therefore, answer is Ca.

Q17. At RTP, 1 mol of gas (assume ideal) occupies 24 000 cm3.

mole of O2 = 300 /24 000 = 0.0125 mol

2Ca + O2 --> 2 CaO
2Mg + O2 --> 2 MgO
4K + O2 --> 2K2O
4Na + O2 --> 2 Na2O

0.0125 mol of O2 will react with:
0.025 mol of Ca (1 g)
0.05 mol of Mg (1.2 g)
0.05 mol of K (1.95 g)
0.05 mol of Na (1.15 g)
Click to expand...


Thanks a lot! I appreciate all ur help :)
 
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