Chemistry: Post your doubts here!

SohaibButt · May 29, 2017

Anyone?

Dukula Jayasinghe · May 29, 2017

SohaibButt said:
Anyone?

Oh this one. See, now when these acids react with calcium, there are several structural formulae. So for ethanoic acid it will be (CH3COO)2Ca. This will be C4H6O4Ca. Now this doesnt match up with the answer. So 1 is incorrect. 2 & 3 are hence correct. This is not AS question. This is a higher level one. But cambridge have given it in such a way that the AS candidate can cross out the first answer to get to the correct answer . Tactics. The other two molecules are complex, so not to worry.

Minahil rizvi · May 29, 2017

techgeek said:
When ethane reacts with Bromine, it forms CH2BrCH3 but the product can react with Br2 again and form disubstituted product which reacts again with Br2 and form trisbustituted product and so on, we can actually form even hexasubstituted product. So it goes like this:
Monosubstituted:
CH2BrCH3
Disubstituted:
CH2BrCH2Br
and
CHBr2CH3
Trisubstituted:
CH3CBr3
and
CHBr2CH2Br
Tetrasubstituted:
CBr3CH2Br
and
CHbr2CHBr2
Pentasubstituted:
CBr3CHBr2
Hexasubstituted:
CBr3CBr3

yesss, thankyouuu

amina1300 · May 30, 2017

Can anyone help me with these : http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_qp_13.pdf

[Q7] Use of the Data Booklet is relevant for this question. In an experiment, the burning of 1.45g (0.025mol) of propanone was used to heat 100g of water. The initial temperature of the water was 20.0°C and the final temperature of the water was 78.0°C. Which experimental value for the enthalpy change of combustion for propanone can be calculated from these results?
A –1304kJ mol–1 B –970kJ mol–1 C –352kJ mol–1 D –24.2kJ mol–1

[Q9] Please.

Q12 X is a Group II metal. The carbonate of X decomposes when heated in a Bunsen flame to give carbon dioxide and a white solid residue as the only products. This white solid residue is sparingly soluble in water. Even when large amounts of the solid residue are added to water the pH of the saturated solution is less than that of limewater. What could be the identity of X?
A magnesium B calcium C strontium D barium

Q21 please, I dont get how to solve this one.

Q24 This one too.

Q38

I'll be so happy if these questions are solved

amina1300 · May 30, 2017

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_qp_12.pdf

4
Flask X contains 5dm3 of helium at 12kPa pressure and flask Y contains 10dm3 of neon at 6kPa pressure. If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9kPa C 10kPa D 11kPa

amina1300 · May 30, 2017

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_13.pdf

10 Use of the Data Booklet is relevant to this question. 1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0dm3 .
What could be the identity of the metal?
A calcium B magnesium C potassium D sodium

Q22.

26 Alkane X has molecular formula C4H10. X reacts with Cl 2(g) in the presence of sunlight to produce only two different monochloroalkanes, C4H9Cl. Both of these monochloroalkanes produce the same alkene Y, and no other organic products, when they are treated with hot ethanolic KOH.
What is produced when Y is treated with hot concentrated acidified KMnO4?
A CO2 and CH3CH2CO2H B CO2 and CH3COCH3 C HCO2H and CH3COCH3 D CH3CO2H only

28 This one is really confusing.

Dukula Jayasinghe · May 30, 2017

amina1300 said:
Can anyone help me with these : http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_qp_13.pdf

[Q7] Use of the Data Booklet is relevant for this question. In an experiment, the burning of 1.45g (0.025mol) of propanone was used to heat 100g of water. The initial temperature of the water was 20.0°C and the final temperature of the water was 78.0°C. Which experimental value for the enthalpy change of combustion for propanone can be calculated from these results?
A –1304kJ mol–1 B –970kJ mol–1 C –352kJ mol–1 D –24.2kJ mol–1

[Q9] Please.

Q12 X is a Group II metal. The carbonate of X decomposes when heated in a Bunsen flame to give carbon dioxide and a white solid residue as the only products. This white solid residue is sparingly soluble in water. Even when large amounts of the solid residue are added to water the pH of the saturated solution is less than that of limewater. What could be the identity of X?
A magnesium B calcium C strontium D barium

Q21 please, I dont get how to solve this one.

Q24 This one too.

Q38

I'll be so happy if these questions are solved

(Q7) E=mc(Temperature change)
assuming 1g ~ 1cm3
E = 100 x 4.18 x (78-20)
E= 24244 J
Now since propanone is limiting u take its moles into consideration here
Change in enthalpy = -(E/0.025) = -969.76 KJmol-1

(Q8)
Down the group, thermal stability of group II carbonates increases and solubility increases.
As solubility increases, the value of the pH decreases
Ba and Sr is both stronger than Ca and more soluble so not C and D.
Ca(OH)2 is lime water so not B. Hence, answer is A.

(Q21)
Oh. this is a bit hard question. U need to think a lot. First, in my opinion, draw the displayed formulae for each compound provided. Eliminate B as it contains only 13 carbon atoms but Q contains 14 carbon atoms. Take (A) into consideration first. This compound only has the corner branch of Q twisted down so it is not an isomer of Q. Take (D) into consideration. That has the same thing as (A). Displayed formulae miight help u with this. Hence (C) is the answer.

(Q24)
SN1 Nucleophillic substitution reactions is undergone by tertiary halagenoalkanes. (3 carbons bonded to carbon containing halogen). (C) only clearly shows this fact. Hence (C) is the answer.

(Q9)
Eliminate (C) and (D) both cos they are not empirical formulae.
Number of Carbon atoms - 6
Number of Hydrogen atoms - 12
Number of Nitrogen atoms - 4
Divide everything by 2 to get empirical formula.
It comes to C3H6N2 (B)

(Q38)
3 gives only 1 product. This is because when the OH goes away from the methyl group a C=C bond is formed with the cyclo compound with the methyl. Hence a double bond cannot be formed with the carbon-carbon bond joined to the methyl group. Hope u get the idea. It's hard to explain. Look at the molecule closely. 2 is obviousy correct as either side the C=C bond goes to, it is the same formula. 1 is wrong cos when u put C=C bond to each side, when counting to get structural formula u see it is different.
Thanks

Dukula Jayasinghe · May 30, 2017

amina1300 said:
http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_qp_12.pdf

4
Flask X contains 5dm3 of helium at 12kPa pressure and flask Y contains 10dm3 of neon at 6kPa pressure. If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9kPa C 10kPa D 11kPa

The answer is 8 kPa.
This is a bit hard question. Use of PV=nRT is necessary.
First calculate number of Heilium moles initialy. Use PV=nRT.
PV=nRT
12 x 1000 x 5 x 0.001 = n(He) x R x T
n (He) = (12 x 5) / (RT)
Then calculate number of Neon moles initially.
PV=nRT
6 x 1000 x 10 x 0.001 = n(Ne) x R x T
n (Ne) = (6 x 10) / (RT)
Then now u must calculate total final pressure
PV = nRT
PV = (n(He)+n(Ne)) x R x T
P (5+10) = [(60/RT) + (60/RT)] x R x T
p (15 x 0.001) = 120/RT x RT
P=120 / (15 x 0.001)
P = 8000 Pa
P = 8 kPa

Dukula Jayasinghe · May 30, 2017

amina1300 said:
http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_13.pdf

10 Use of the Data Booklet is relevant to this question. 1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0dm3 .
What could be the identity of the metal?
A calcium B magnesium C potassium D sodium

Q22.

26 Alkane X has molecular formula C4H10. X reacts with Cl 2(g) in the presence of sunlight to produce only two different monochloroalkanes, C4H9Cl. Both of these monochloroalkanes produce the same alkene Y, and no other organic products, when they are treated with hot ethanolic KOH.
What is produced when Y is treated with hot concentrated acidified KMnO4?
A CO2 and CH3CH2CO2H B CO2 and CH3COCH3 C HCO2H and CH3COCH3 D CH3CO2H only

28 This one is really confusing.

(Q10) Moles of O2 = 300 / 24000 = 0.0125 mol
Now find the mass of O2 = 0.0125 x 32 = 0.4g
Emprical formulae method
X : O
1g : 0.4g / 16
1g / Mr(Y) : 0.025
You know Ca, Mg, K and Na forms oxides with CaO, MgO, K2O, Na2O
For Ca and Mg
the same number of moles as of O2 should be present to make ratio 1:1
Hence; 1g / Mr = 0.025 1 / 0.025 = Mr Mr = 40 g mol-1. Hence Ca = Mr = 40 = (A)
For K and Na, twice the number of moles of O2 should be present to make ratio 2:1
Hence ; 1/0.050 = 20 g mol. (C) , (D) are incorrect.
THERE ARE EASIER DIFFERENT APPROACHES TO THIS METHOD

(q22)
U have to draw all the displayed formula for (A) to (D) and for the compound provided. You can see that when u replace the halogens with Iodine, only D matches with that of the compound given.

(Q26)
You gotta combine everything first to make the initial compound.
For (A)
CH3CH2CO2H + CO2 --> CH3CH2CH=CH2 --> CH3CH2CHClCH3 or CH3CH2CH2CH2Cl. Hence as the final compounds when reacted to eliminate the halogens, both will not give same products. A is eliminated.
For (B)
CH3COCH3 + CO2 --> (CH3)2C=CH2 is the only possible product. Hence when chlorine is added, it will become --> (CH3)2CCH2Cl, but in anyway you look at the molecule the naming will be the same. (Draw displayed formula and see). Hence (B) is the answer.
(C) is incorrect as methanoic acid is formed which cannot be formed as when >C=CH2 bond breaks CH2 forms CO2.
(D) is totally incorrect as CH3CO2H --> CH3CH=CHCH3 --> CH3CHClCHCH3 or CH3CH2CH2CH2Cl can be both formed.

(Q28)
This is a SN1 type reaction. In SN1 type reactions only the organic compound and the new replacing nucleophile is in the intermediate. The first step is the heterolytic bond clevage of the C-Br bond. Therefore a C+ (carbocation) is formed in the intermediate where it is attacked by the OH nucleophile since it is highly unstable. The activation energy required to break the C-Br bond is higher than the energy required to form the C-OH bond. First step is endothermic since bond is broken and next step is exothermic since bonds are made. Hence the diagram is given as that confusing graph. So in the middle , it is the intermediate with a carbocation so the answer is C.

amina1300 · May 30, 2017

Dukula Jayasinghe said:
(Q10) Moles of O2 = 300 / 24000 = 0.0125 mol
Now find the mass of O2 = 0.0125 x 32 = 0.4g
Emprical formulae method
X : O
1g : 0.4g / 16
1g / Mr(Y) : 0.025
You know Ca, Mg, K and Na forms oxides with CaO, MgO, K2O, Na2O
For Ca and Mg
the same number of moles as of O2 should be present to make ratio 1:1
Hence; 1g / Mr = 0.025 1 / 0.025 = Mr Mr = 40 g mol-1. Hence Ca = Mr = 40 = (A)
For K and Na, twice the number of moles of O2 should be present to make ratio 2:1
Hence ; 1/0.050 = 20 g mol. (C) , (D) are incorrect.
THERE ARE EASIER DIFFERENT APPROACHES TO THIS METHOD

(q22)
U have to draw all the displayed formula for (A) to (D) and for the compound provided. You can see that when u replace the halogens with Iodine, only D matches with that of the compound given.

(Q26)
You gotta combine everything first to make the initial compound.
For (A)
CH3CH2CO2H + CO2 --> CH3CH2CH=CH2 --> CH3CH2CHClCH3 or CH3CH2CH2CH2Cl. Hence as the final compounds when reacted to eliminate the halogens, both will not give same products. A is eliminated.
For (B)
CH3COCH3 + CO2 --> (CH3)2C=CH2 is the only possible product. Hence when chlorine is added, it will become --> (CH3)2CCH2Cl, but in anyway you look at the molecule the naming will be the same. (Draw displayed formula and see). Hence (B) is the answer.
(C) is incorrect as methanoic acid is formed which cannot be formed as when >C=CH2 bond breaks CH2 forms CO2.
(D) is totally incorrect as CH3CO2H --> CH3CH=CHCH3 --> CH3CHClCHCH3 or CH3CH2CH2CH2Cl can be both formed.

(Q28)
This is a SN1 type reaction. In SN1 type reactions only the organic compound and the new replacing nucleophile is in the intermediate. The first step is the heterolytic bond clevage of the C-Br bond. Therefore a C+ (carbocation) is formed in the intermediate where it is attacked by the OH nucleophile since it is highly unstable. The activation energy required to break the C-Br bond is higher than the energy required to form the C-OH bond. First step is endothermic since bond is broken and next step is exothermic since bonds are made. Hence the diagram is given as that confusing graph. So in the middle , it is the intermediate with a carbocation so the answer is C.

OMG THANKYOUUUUUUU SOO MUUCCCHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Holmes · May 30, 2017

wolfestein99x said:
Hey yo sherlock the answer is A , ya didnt help much XD

Oh sorry

bytheway thanks for telling my mistake.
Good luck

sara_queen · May 31, 2017

https://www.slideshare.net/fahadhameedahmad1/as-practical-exam-notes

for measuring mass why dont we minus the empty bottle from the mass of empty bottle plus FA2 to find the volume of FA2?

selrey · May 31, 2017

Someone help me with this qs
X and Y are oxides of different Period 3 elements. If one mole of Y is added to water, the solution formed is neutralised by exactly one mole of X. What could be the identities of X and Y?
X Y
A Al 2O3 P4O10
B Al 2O3 SO3
C Na2O P4O10
D Na2O SO3

amina1300 · May 31, 2017

selrey said:
Someone help me with this qs
X and Y are oxides of different Period 3 elements. If one mole of Y is added to water, the solution formed is neutralised by exactly one mole of X. What could be the identities of X and Y?
X Y
A Al 2O3 P4O10
B Al 2O3 SO3
C Na2O P4O10
D Na2O SO3

OK so first you need to know the reactions of the period 3 oxides with water. AL2O3 does not react so optionA b are wrong. Next we react 2NaOH (which is a basic oxide and the product of reaction between oxide and H2O) with corresponding acidic oxide & balance the equations.
NOTE The Q states"neutralised by exactly 1mol....so D is correct as it balances perfectly with 2NaOH.

amina1300 · May 31, 2017

POST THE HARDEST MCQs!

amina1300 · May 31, 2017

Dukula Jayasinghe said:
(Q10) Moles of O2 = 300 / 24000 = 0.0125 mol
Now find the mass of O2 = 0.0125 x 32 = 0.4g
Emprical formulae method
X : O
1g : 0.4g / 16
1g / Mr(Y) : 0.025
You know Ca, Mg, K and Na forms oxides with CaO, MgO, K2O, Na2O
For Ca and Mg
the same number of moles as of O2 should be present to make ratio 1:1
Hence; 1g / Mr = 0.025 1 / 0.025 = Mr Mr = 40 g mol-1. Hence Ca = Mr = 40 = (A)
For K and Na, twice the number of moles of O2 should be present to make ratio 2:1
Hence ; 1/0.050 = 20 g mol. (C) , (D) are incorrect.
THERE ARE EASIER DIFFERENT APPROACHES TO THIS METHOD

(q22)
U have to draw all the displayed formula for (A) to (D) and for the compound provided. You can see that when u replace the halogens with Iodine, only D matches with that of the compound given.

(Q26)
You gotta combine everything first to make the initial compound.
For (A)
CH3CH2CO2H + CO2 --> CH3CH2CH=CH2 --> CH3CH2CHClCH3 or CH3CH2CH2CH2Cl. Hence as the final compounds when reacted to eliminate the halogens, both will not give same products. A is eliminated.
For (B)
CH3COCH3 + CO2 --> (CH3)2C=CH2 is the only possible product. Hence when chlorine is added, it will become --> (CH3)2CCH2Cl, but in anyway you look at the molecule the naming will be the same. (Draw displayed formula and see). Hence (B) is the answer.
(C) is incorrect as methanoic acid is formed which cannot be formed as when >C=CH2 bond breaks CH2 forms CO2.
(D) is totally incorrect as CH3CO2H --> CH3CH=CHCH3 --> CH3CHClCHCH3 or CH3CH2CH2CH2Cl can be both formed.

(Q28)
This is a SN1 type reaction. In SN1 type reactions only the organic compound and the new replacing nucleophile is in the intermediate. The first step is the heterolytic bond clevage of the C-Br bond. Therefore a C+ (carbocation) is formed in the intermediate where it is attacked by the OH nucleophile since it is highly unstable. The activation energy required to break the C-Br bond is higher than the energy required to form the C-OH bond. First step is endothermic since bond is broken and next step is exothermic since bonds are made. Hence the diagram is given as that confusing graph. So in the middle , it is the intermediate with a carbocation so the answer is C.

Can u plz explain Q 26 a bit more? I dont get how Oxygen "CH3CH2CO2H + CO2 --> CH3CH2CH=CH2 -" disappears and how 1 or 2 products form.

Dukula Jayasinghe · May 31, 2017

amina1300 said:
Can u plz explain Q 26 a bit more? I dont get how Oxygen "CH3CH2CO2H + CO2 --> CH3CH2CH=CH2 -" disappears and how 1 or 2 products form.

Send me your whatsapp number in private. Ill draw and show u.

SohaibButt · May 31, 2017

Can i post biology question over here? xD

Dukula Jayasinghe · May 31, 2017

d

SohaibButt said:
Can i post biology question over here? xD

dONT Man. It will be utterly useless.

amina1300 · May 31, 2017

Does anyone have complete ORGANIC notes I lost mine

I (need them for the MCQ exam because I forgot most of the reactions and important points)

I will be really grateful if someone shares their...

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