P1 MCQ's preparation thread for chemistry ONLY!!!!

[Raweeha]

Okay, since it's bond energy, it's always +ve, energy is taken in to break the bond, that eliminates C ... now X-Y should be broken completely to form a certain amount of only X and a certain amount of only Y. That eliminates B. Now A and D. In D, the bond Y-Y is reformed, do that does not give us the exact enthalpy of X-Y as energy is released. Which leaved A, in which a whole no. of moles of only X and only Y

 

[Raweeha]

no problem!! plz see the other question posted.

Done!

 

[HorsePower]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Q1

 

[itallion stallion]

Okay, since it's bond energy, it's always +ve, energy is taken in to break the bond, that eliminates C ... now X-Y should be broken completely to form a certain amount of only X and a certain amount of only Y. That eliminates B. Now A and D. In D, the bond Y-Y is reformed, do that does not give us the exact enthalpy of X-Y as energy is released. Which leaved A, in which a whole no. of moles of only X and only Y

thanks alot!!!!!

 

[Raweeha]

thanks alot!!!!!

Anytime

 

[samiaaaaaaaaaaaaahhh]

:O half the paper :O well relx first take a deep breath and reattempt it you wil do it

these are 9 questions not half of the paper

 

[Razo513]

can someone please explain this

 

[Razo513]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Q1

the answer is A .. because as u increase temperature particles with energy higher than the activation energy increased .... keep i mind that the area under the line represents the particles with 3 levels of energy low, mid and high ... thus the particles at the high side that is the end of the graph should increase ...so the solid line should be below the dotted line

 

[Obsidian Fl1ght]

In the first reaction SO2 reduces Cl2 to HCL. So it is stronger reducing agent than Cl2. SO2 > Cl2
In the second reaction H2S reduces Cl2 to HCl. So it is stronger than Cl2. H2S > Cl2
In third reaction H2S reduces SO2 to S. So it's stronger than SO2. H2S> SO2
H2S> SO2> Cl2

Thank u loads! I'd been driving myself crazy trying to do that question.

First one is a tertiary alcohol. If you draw out the structure you would see that it cannot be dehydrated as the carbon adjacent to the carbon which is attached to OH group has no H attached. The second one can be dehydrated using Sulpuric Acid.

Okay got it. Thnx again.

 

[Obsidian Fl1ght]

View attachment 28790

can someone please explain this

K, so we gotta have a Precipitate by addition of Ca(OH)2.
The easiest, (And only one that I can think of now!), way is to look at the choices and decide which MAY form ppts. Aka which salts are insoluble and form ppts.

C and D are outta the game.
Why?
Coz they''re nitrate and ALL nitrates are soluble. Mg(NO3)2 and Ca(NO3)2 will not form any ppts.

From A and B:
A: All chlorides are soluble except silver, lead, mercury chlorides.
So CaCl2 will be soluble = no ppt.

C: All carbonates are insoluble except Grp I and Ammonium (Carbonates).
So CaCO3 will form a ppt.

Ans: C.

 

[abdullah100]

This is a tough question i need help.
9701_s12_qp_12.pdf Q7
9701_w08_qp_1.pdf Q30

 

[queen of the legend]

Given the following enthalpy changes,
I2(g) + 3Cl2(g) → 2ICl3(s) ΔHo = –214 kJ mol–1
I2(s) → I2(g) ΔHo = +38 kJ mol–1
What is the standard enthalpy change of formation of iodine trichloride, ICl3(s)?
A +176 kJ mol–1
B –88 kJ mol–1
C –176 kJ mol–1
D –214 kJ mol–1

ans is B ...can sm1 expalin plss

 

[Eshika]

Given the following enthalpy changes,
I2(g) + 3Cl2(g) → 2ICl3(s) ΔHo = –214 kJ mol–1
I2(s) → I2(g) ΔHo = +38 kJ mol–1
What is the standard enthalpy change of formation of iodine trichloride, ICl3(s)?
A +176 kJ mol–1
B –88 kJ mol–1
C –176 kJ mol–1
D –214 kJ mol–1

ans is B ...can sm1 expalin plss

38 - 214 = -176 and -176/2 = -88
you divide by 2 becoz there are 2 iodine molecules present

 

[queen of the legend]

10 The value of the equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol
and ethanoic acid is 4.0 at 60 °C.
C2H5OH + CH3CO2H CH3CO2C2H5 + H2O
When 1.0 mol of ethanol and 1.0 mol of ethanoic acid are allowed to reach equilibrium at 60 °C,
what is the number of moles of ethyl ethanoate formed?

i got the answer as 0.4 or 2/5 ...but MS its 0.6 or 2/3!

 

[daisyy]

hey can anyone please help me in Question number 20 http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf and I also need help in question number 12 Oct/Nov 2005 Thanks guys!

 

[daisyy]

10 The value of the equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol
and ethanoic acid is 4.0 at 60 °C.
C2H5OH + CH3CO2H CH3CO2C2H5 + H2O
When 1.0 mol of ethanol and 1.0 mol of ethanoic acid are allowed to reach equilibrium at 60 °C,
what is the number of moles of ethyl ethanoate formed?

i got the answer as 0.4 or 2/5 ...but MS its 0.6 or 2/3!

C2H5OH+CH3CO2H --> CH3COOC2H5 + H2O
at start 1 1 1 0 0
amount reacted x x x x
At equilibrium 1-x 1-x x x
therefore Kc= x^2/(x-1)^2 = 4
cross multiply and u'll get the right answer! Can you help me with the question I posted please? Thanks

 

[Razo513]

hey can anyone please help me in Question number 20 http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf and I also need help in question number 12 Oct/Nov 2005 Thanks guys!

Q20 i cant think of any reasonable explanation for this question other than the OH- being a nucleophile

as for Q12 ... they said 1 mole of Mg , Al and S so write down all 3 combustion equations and balance but keeping 1 mole of those 3 .. 1) Mg + 1/2 o2--> MgO 2) Al +3/4O2--> Al2O3 3) S+3/2O2-->SO3 notice the moles of O2 in those combustion eqns are increasing but not equally thus the only graph representing this is D

 

[Mairaxo]

Sorry for Bro #examtime
What do you mean that an O is attached to both end of alkene? CAn you tell me the product of this reation?

if an alkene reacts with cold KMnO4 the double bond breaks nd -OH groups are attached to both ends. For example, CH2=CH2 will become HO-CH2-CH2-OH
if hot acidic KMnO4 is used O is added and further oxidation makes an acid. For example. CH2=CH2 becomes 2 mol CH2O. CH2O will get oxidised to CO2 and H2O.
Example 2- CH3-CH=CH-CH3 will become 2 mol CH3-CHO and further oxidation will make CH3COOH.
Example 3- (CH3)2C=CH2 will become (CH3)2CO and CH2O and further oxidation will produce CO2 and H2O from CH2O whereas (CH3)2CO will not get oxidised as its a ketone

 

[queen of the legend]

hey can anyone please help me in Question number 20 http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf and I also need help in question number 12 Oct/Nov 2005 Thanks guys!

for the 2006 question : i did not understand it myself !
but the 2005 question 12 its very simple: make all no. of moles of elements equal or 2 (just to compare easily)

2 Mg + O2 --------> 2 MgO

2Al + 1.5 O2 ---------> Al2O3

2 S + 2O2 ----------> 2SO2

look at the no. of moles of oxygen used in each equation and decide !

 

[daisyy]

for the 2006 question : i did not understand it myself !
but the 2005 question 12 its very simple: make all no. of moles of elements equal or 2 (just to compare easily)

2 Mg + O2 --------> 2 MgO

2Al + 1.5 O2 ---------> Al2O3

2 S + 2O2 ----------> 2SO2

look at the no. of moles of oxygen used in each equation and decide !

Lool! How cool! Okay umm in the nov 05 question in this case isnt the answer supposed to be A? Cuz the difference between them is constant 0.5...