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Chemistry: Post your doubts here!

Messages
98
Reaction score
80
Points
28
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
 
Messages
98
Reaction score
80
Points
28
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
 
Messages
98
Reaction score
80
Points
28
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.
 
Messages
363
Reaction score
194
Points
53
(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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 
Messages
50
Reaction score
24
Points
18
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
 
Messages
363
Reaction score
194
Points
53
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.
 
Last edited:
Messages
363
Reaction score
194
Points
53
(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.
 
Messages
363
Reaction score
194
Points
53
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...
 
Messages
150
Reaction score
222
Points
53
Could anyone help me with this. Looks easy but my results don't match!!!!!

12 Use of the Data Booklet is relevant to this question.
This question should be answered using bond enthalpy data. The equation for the complete combustion of methane is given below.
CH4 + 2O2 → CO2 + 2H2O
What is the enthalpy change of combustion of methane?
A –1530kJ mol–1
B –1184kJ mol–1
C –770kJ mol–1
D –688kJ mol–1
 
Messages
438
Reaction score
3,645
Points
253
Could anyone help me with this. Looks easy but my results don't match!!!!!

12 Use of the Data Booklet is relevant to this question.
This question should be answered using bond enthalpy data. The equation for the complete combustion of methane is given below.
CH4 + 2O2 → CO2 + 2H2O
What is the enthalpy change of combustion of methane?
A –1530kJ mol–1
B –1184kJ mol–1
C –770kJ mol–1
D –688kJ mol–1

To my inadequate knowledge the answer is "D"
CH4 + 2O2 → CO2 + 2H2O

Bond energy of reactants:
=4(410)+2(496)
= +2632 kj/mol

now look at The bond energy of Products:
= 2 (740)+ 4 (460)
= +3320 kj/mol

last step:
enthalpy change = B.E of Reactants - B.E. of Products
= 2632 -3320
= - 688 kj/mol

HOPE IT HELPS! :D
 
Messages
150
Reaction score
222
Points
53
To my inadequate knowledge the answer is "D"
CH4 + 2O2 → CO2 + 2H2O

Bond energy of reactants:
=4(410)+2(496)
= +2632 kj/mol

now look at The bond energy of Products:
= 2 (740)+ 4 (460)
= +3320 kj/mol

last step:
enthalpy change = B.E of Reactants - B.E. of Products
= 2632 -3320
= - 688 kj/mol

HOPE IT HELPS! :D
Your answer is correct, but if you look in the Data Booklet, In the Bond Energies section in polyatomic molecules section: there are different values for C double bond C and C Double Bond C in CO2.
 
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