Chemistry: Post your doubts here!

[kanchan dhiman]

solved paper 1 with reasoning

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
Q3 part (c)
Why is it Br2 for MgBr2?

 

[Nidz-Ahmed]

The temperature at which the Atmospheric pressure is equal to the vapor pressure is the boiling point of the liquid .... can Someone please Explain the Concept Behind this ?

 

[daredevil]

When bond energy is low then standard enthalpy change of combistion is high??

 

[Nidz-Ahmed]

When bond energy is low then standard enthalpy change of combistion is high??

Standard Enthalpy is high When large Amount of energy is released , So When the bond energy is low , the bonds are easily dissociated , so when products Are formed large amount of energy is released to the surrounding

 

[Nidz-Ahmed]

When bond energy is low then standard enthalpy change of combistion is high??

Uhm i Dunt knw that was just a guess

 

[Warrior66]

Please help with question no. 5, 10, 12, 21

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf

Question 5:
20% of 10g is 2g. now use formula (moles=mass/Mr) ---> 2g/59 (you'll need to look this up on the data booklet) = 0.0339 mol
you must know that Avogadro's number= 6.02 x 10^23 so ----> (6.02 x 10^23) x 0.0339 = 2.04 x 10^22 which gives the answer as A.
Question 10:
Kc = [CH3CO2C2H5] x [H2O]/[C2H5OH] x [CH3CO2H]

C2H5OH + CH3CO2H ------>CH3CO2C2H5 + H2O

initial conc. 1 1 0 0
equilibrium 1-x 1-x x x
conc.
(^ the values keep coming together. note
that each value goes for each reactant/product.
square root of 4.0 = (x) / (1-x)
:. 2(1-x) = x
2-2x = x
2 = x + 2x
2 = 3x
x = 2/3
final answer B.
Question 12:
calculate the Mr's of each option. (although it isn't necessary because just by looking at option C you realise that Mr of H= 16 x 1= 16 and Mr of O= 1 x 16 = 16 because there is only 1 O in the compound) so whatever the total mass is of the compound dividing the mass of O or H by the total mass and multiplying it with 100 to get the %age...will give you the same answer for both.
Question 21:
look at the ester, it is called (Esters names have two words, the first word comes from the alcohol
portion, and the second word is derived from the acid portion.) butyl ethanoate
the -O- is the ester bond and it is formed when the alcohol and carboxylic acid combine in a condensation reaction- water is lost in this reaction.
when you add water back to this compound the -OH- goes back to its alcohol so count how many C's there are in the alcohol on the left side of the -O- bond and how many H's there are PLUS the one from -OH- and how many O's there are PLUS the one from -OH- ----> final answer C4H8O2
D.

 

[unseen95]

please help me with question number 2(a)(i) with explanation http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_41.pdf

 

[Raiyan3]

Question 5:
20% of 10g is 2g. now use formula (moles=mass/Mr) ---> 2g/59 (you'll need to look this up on the data booklet) = 0.0339 mol
you must know that Avogadro's number= 6.02 x 10^23 so ----> (6.02 x 10^23) x 0.0339 = 2.04 x 10^22 which gives the answer as A.
Question 10:
Kc = [CH3CO2C2H5] x [H2O]/[C2H5OH] x [CH3CO2H]

C2H5OH + CH3CO2H ------>CH3CO2C2H5 + H2O

initial conc. 1 1 0 0
equilibrium 1-x 1-x x x
conc.
(^ the values keep coming together. note
that each value goes for each reactant/product.
square root of 4.0 = (x) / (1-x)
:. 2(1-x) = x
2-2x = x
2 = x + 2x
2 = 3x
x = 2/3
final answer B.
Question 12:
calculate the Mr's of each option. (although it isn't necessary because just by looking at option C you realise that Mr of H= 16 x 1= 16 and Mr of O= 1 x 16 = 16 because there is only 1 O in the compound) so whatever the total mass is of the compound dividing the mass of O or H by the total mass and multiplying it with 100 to get the %age...will give you the same answer for both.
Question 21:
look at the ester, it is called (Esters names have two words, the first word comes from the alcohol
portion, and the second word is derived from the acid portion.) butyl ethanoate
the -O- is the ester bond and it is formed when the alcohol and carboxylic acid combine in a condensation reaction- water is lost in this reaction.
when you add water back to this compound the -OH- goes back to its alcohol so count how many C's there are in the alcohol on the left side of the -O- bond and how many H's there are PLUS the one from -OH- and how many O's there are PLUS the one from -OH- ----> final answer C4H8O2
D.

Thanks thanks alot bro! Now i get it! Your help was greatly appreciated!

 

[taimoor.08]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w02_qp_1.pdf Q1

 

[Gémeaux]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w02_qp_1.pdf Q1

Write equations for each reaction,
CH4 + 2O2 --> CO2 + H2O and C2H6 + 7/2 O2 --> 2CO2 + 3H2O
In first equation, 1 mole of methane produces one mole of CO2, and as 1 mole of any gas occupies 24dm^3 at rtp, 10 cm^3 of methane produces 10cm^3 CO2.
In second reaction, one mole of ethane burns to give Two mole carbondioxide, hence 10cm^3 gives 20cm^3 of gas.
KOH absorbs CO2, hence total volume it absorbs is the total volume of CO2 produced, that is 10 + 20 = 30cm^3. Hence, C is the correct option.

 

[..sacrifice4Revenge..]

any resources for organic chem here..? :S

 

[iKhaled]

please help me with question number 2(a)(i) with explanation http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_41.pdf

ok, a mathematical way to answer this question is to look at 2 experiments in the table which has one thing different only then divide them over each other for example..

experiment 2 and 3..we will divide the rate equation of the 2 by each other. first of all the rate equation in general is r= k [CH3CHO]^x [CH3OH]^y [H+]^z

1.25 = [0.25]^x [0.10]^y [0.05]^z
2 .00= ----------------------------------
[0.25]^x [0.16]^y [0.05]^z

do the math and u will get that 0.625 = 0.625^y which means y is 1...if we go back to our rate equation we will see that y is the order of respect of CH3OH so the order with respect to CH3OH is one

now pick up 2 other experiments and apply the same thing. let's look at experiment one and 2

1 = [0.20]^x [0.10]^1 [0.05]^z
1.25= ------------------------------
[0.25]^x [0.10]^1 [0.05]^z

do the division again and u will get 0.8 = 0.8^x which means that it x = 1 so its with respect to one with CH3CHO.

do the same thing to find z which will tell us the order with respect to H+

i hope u got it..its easier to explain this ftf btw :/

 

[unseen95]

ok, a mathematical way to answer this question is to look at 2 experiments in the table which has one thing different only then divide them over each other for example..

experiment 2 and 3..we will divide the rate equation of the 2 by each other. first of all the rate equation in general is r= k [CH3CHO]^x [CH3OH]^y [H+]^z

1.25 = [0.25]^x [0.10]^y [0.05]^z
2 .00= ----------------------------------
[0.25]^x [0.16]^y [0.05]^z

do the math and u will get that 0.625 = 0.625^y which means y is 1...if we go back to our rate equation we will see that y is the order of respect of CH3OH so the order with respect to CH3OH is one

now pick up 2 other experiments and apply the same thing. let's look at experiment one and 2

1 = [0.20]^x [0.10]^1 [0.05]^z
1.25= ------------------------------
[0.25]^x [0.10]^1 [0.05]^z

do the division again and u will get 0.8 = 0.8^x which means that it x = 1 so its with respect to one with CH3CHO.

do the same thing to find z which will tell us the order with respect to H+

i hope u got it..its easier to explain this ftf btw :/

can we take any equation or is there a compulsion of taking a specific one.

 

[unseen95]

ok, a mathematical way to answer this question is to look at 2 experiments in the table which has one thing different only then divide them over each other for example..

experiment 2 and 3..we will divide the rate equation of the 2 by each other. first of all the rate equation in general is r= k [CH3CHO]^x [CH3OH]^y [H+]^z

1.25 = [0.25]^x [0.10]^y [0.05]^z
2 .00= ----------------------------------
[0.25]^x [0.16]^y [0.05]^z

do the math and u will get that 0.625 = 0.625^y which means y is 1...if we go back to our rate equation we will see that y is the order of respect of CH3OH so the order with respect to CH3OH is one

now pick up 2 other experiments and apply the same thing. let's look at experiment one and 2

1 = [0.20]^x [0.10]^1 [0.05]^z
1.25= ------------------------------
[0.25]^x [0.10]^1 [0.05]^z

do the division again and u will get 0.8 = 0.8^x which means that it x = 1 so its with respect to one with CH3CHO.

do the same thing to find z which will tell us the order with respect to H+

i hope u got it..its easier to explain this ftf btw :/

thank you very much for helping. I appreciate it so much

 

[snoonono]

Can someone explain Hess' law real fast and easy way? Thanks

 

[xxfarhaxx]

cn sum1 pls hlp!! -- Q4 , c (ii)
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_42.pdf
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_ms_42.pdf
why do we divide the ksp value by 4??

 

[Tkp]

cn sum1 pls hlp!! -- Q4 , c (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_42.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_ms_42.pdf
why do we divide the ksp value by 4??

CONC OF pb is x and cl2 is 2x
so x*2x^2=2.0 × 10–5

 

[lyuchenqujoy]

Yeah.. exactly! I asked my teacher the same question, but she said that it needs knowledge which is beyond A levels
So just remember:
When the forward reaction is exothermic, increasing the temperature decreases the value of the equilibrium constant.
When the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant.

all right, thank you very muchhhhhhhhhh anyway XD!!!

 

[xxfarhaxx]

CONC OF pb is x and cl2 is 2x
so x*2x^2=2.0 × 10–5

tnx