furuta

hello~
can someone please solve the following problem for me?
1. Use of the Data Booklet is relevant to this question.
The gas laws can be summarised in the ideal gas equation.

pV = nRT
0.96 g of oxygen gas is contained in a glass vessel of volume 7000 cm3 at a temperature of 30 °C.

What is the pressure in the vessel?
A 1.1kPa B 2.1kPa C 10.8kPa D 21.6kPa

thank you x

SohaibButt

hello~
can someone please solve the following problem for me?
1. Use of the Data Booklet is relevant to this question.
The gas laws can be summarised in the ideal gas equation.

pV = nRT
0.96 g of oxygen gas is contained in a glass vessel of volume 7000 cm3 at a temperature of 30 °C.

What is the pressure in the vessel?
A 1.1kPa B 2.1kPa C 10.8kPa D 21.6kPa

thank you x
P×7000×10*-6=0.96/32×8.31×303

Ans:10.8 kPa

So it's C

Last edited:

shahzaib ihsan

all 3 r correct i think
my question still remains unsolved as i need a detailed explanation regarding why all 3 are correct..

furuta

P×7000×10*-6=0.96/32×8.31×303

Ans:10.8 kPa

So it's C
oh right i accidentally took 16 as the mr instead of 32 alright thank you

shahzaib ihsan

Thank you so much!

Shaaaazxx

Perhaps you can list down the other option(s) you thought were also correct, then we can guide you accordingly.

We can try replacing option A (correct answer) with something familiar

CH4 (g) --> C (g) + 4 H (g) ∆H

Bond energy of C-H is ∆H/4
1/4 CH4 (g) --> 1/4C (g) + H (g)

H4 (g) --> C (g) + 4 H (g) ∆H

Analogously
XYn (g) --> X (g) + n Y (g) ∆H

Bond energy of X-Y is ∆H/n
1/n XYn (g) --> 1/n X (g) + Y (g) ∆H
Thankyou soo muchh!! <33

shahzaib ihsan

K2O + H2SO4 ----> K2SO4 + H2O
Moles of acid = 15/1000*2 = 0.03mol
Moles of K2O in 25cm3 = 0.03mol, due to ratio.
Moles of K2O in 250cm3, = 0.03*10 = 0.3mol
Mass = (39.1*2+16)*0.3 = 28.26g
this question: October/November 2014 variant 12 question number 15.
the equation: K2O + H2SO4 ---> K2SO4 + H2O i don't think this is correct as the question said that K2O is dissolved in water.
so i think these 2 equations may be used to solve the question:
1) K2O + H2O ---> 2KOH
2)
2KOH + H2SO4 ---> K2SO4 + 2H2O.
but i can't seem to find the answer. However i found this on yahoo answers:

(2 mol/dm^3 H2SO4) x (0.015 dm^3) x (2 mol KOH / 1 mol H2SO4) x (1 mol K2O / 2 mol KOH) x
(250 cm^3 / 25 cm^3) x (94.1960 g K2O/mol) = 28 g K2O

can anyone explain this to me on plain paper and/or step by step?

Minahil rizvi

g
How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?
A 3 B 5 C 7 D 9
guyssss help me with this

techgeek

g

guyssss help me with this
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

How does one prove if a reaction is a first order process or not using graph?
What is a first order process?
What are half lives and how do i draw half life lines?
I fear this might come in the practical
Help me out here

Anyone?

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Dukula Jayasinghe

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

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

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

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

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

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

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

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
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