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solved paper 1 with reasoning
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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 surroundingWhen bond energy is low then standard enthalpy change of combistion is high??
Uhm i Dunt knw that was just a guessWhen bond energy is low then standard enthalpy change of combistion is high??
Please help with question no. 5, 10, 12, 21
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.
Write equations for each reaction,
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..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
can we take any equation or is there a compulsion of taking a specific one.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 muchok, 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 :/
CONC OF pb is x and cl2 is 2xcn 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??
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.
tnxCONC OF pb is x and cl2 is 2x
so x*2x^2=2.0 × 10–5
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