Chemistry: Post your doubts here!

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
Q2

Out of every 100 g of fertilizer, 30 g is P205.

Out of 30 g of P2O5 ,

mass of P
= (2 x Mr of P/Mr of P2O5) x 30 g
= (62/142) x 30 g
=13.1 g

 

[ZaqZainab]

Out of every 100 g of fertilizer, 30 g is P205.

Out of 30 g of P2O5 ,

mass of P
= (2 x Mr of P/Mr of P2O5) x 30 g
= (62/142) x 30 g
=13.1 g

Thanks

 

[FarahMJ]

Help please, how do we solve this? Answer is C

 

[ZaqZainab]

Help please, how do we solve this? Answer is C

You have
ΔHc of C2H6 = -1560
ΔHc of H2 = -286
ΔHf of C2H6= -158
Write an equation of formation of C2H6 because the student was trying to calculate ΔHf
C + H2 ---> C2H6
but this isn't balanced
2C+ 3H2---> C2H6
there better
we want ΔHc of C lets take it as x
2x+(3*-286)+1560= -158
+ because C2H6 is in the products and its combustion
make x the subject
x=((-158)-(1560)-(3*-286))/2

 

[Metanoia]

Help please, how do we solve this? Answer is C

Using a cycle, it can look like this.

 

[FarahMJ]

Thank you ZaqZainab and Metanoia so much

 

[FarahMJ]

Can you please help me with these as well

Answers are B and D

 

[ziremm]

I2 (solid) --> I2 (gas)

The bond broken is from one I2 molecule to other I2 molecules, which is induced dipole (as I2 are non polar).

Covalent bonds are NOT broken as the molecules are not broken into atoms.

thank you I thought it was the same atom that was broken..)

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s04_qp_1.pdf
Q9
Q35 isn't 3 suppose to be correct too?
Q37

 

[Youssef Tawil]

Using a cycle, it can look like this.

View attachment 44290

which exam is this?

 

[sadiaali]

Please anyone Q39. I cant understand a word.
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
B is correct.

 

[MariamMalik]

Thank you so so much!!!
You didn't actually get my point though I needed explanation only on 11, 32, 33 and 35.
But anyway thanks so much for taking your time and explaining everything it might be of good use in the end.

Its alright. No problem.

 

[MariamMalik]

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.

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

 

[Metanoia]

Can you please help me with these as well

Answers are B and D

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)

 

[Metanoia]

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

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]

Please anyone Q39. I cant understand a word.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
B is correct.

w08qp1
s13qp13

Two possible radicals could be created from propane.



Radical A reacting with radical B produces structure 1
Radical B reacting with another radical B produces structure 2

 

[Sarah kh]

I dont understand q2 answer is c ,q6 answer is c , q 7 ans is c q9 ans is D andq11 ans is A ,q13 ans A, q14 ans B q15 answer B q16 ans Bnand q20 pls explain atleast some of them
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_11.pdf

 

[UFO_far_planet]

Can someone tell me a good way to draw up the table in question 2: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20%289701%29/9701_w08_qp_31.pdf ?
Thanx in advance!

 

[immie.rose]

Can someone please help me in this question
May 2012 Paper 12 Q23
I would have definitely attached the link but I am from my mobile
Thanks in advance!!!

 

[immie.rose]

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

Thanks!!