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

[afoo.666]

M/J/11/11 q6,8,13.
Can someone pls explain?

 

[Franklin]

Thank you, H4rriet! I like your posts.

 

[itallion stallion]

In MCQ 1 it was written mixture of methane and ethane so i was reacting both of them together with oxygen, in a single equation
In mcq 3, you are right, the nearest is Ni but the ans is Co, why??
And how do we have to do mcq 10, same year, i know all the above 3 options r wrong so that leaves D to be the ans but what if they have given us an option which is considerable, so how do we know by looking at the equation that its exothermic and has large negative enthalpy change of formation??

• In mcq 3 in order to form one mole of an ion with +2 charge we will have to add the first and second ionisation energy.add for both Al and Co have same.if he had said for an ion of +3 charge we will then add first, seond and third.
• in mcq 10 as the reaction is enthlpy change of combustion thus it will be exothermic.so A is wrong and we can never judge a reaction type by its magnitude differance,a reaction can be +1000 or -1000 thus sign matters.so alone 1000 cant give us the detail.

 

[!>@/^~^\@<!]

• In mcq 3 in order to form one mole of an ion with +2 charge we will have to add the first and second ionisation energy.add for both Al and Co have same.if he had said for an ion of +3 charge we will then add first, seond and third.
• in mcq 10 as the reaction is enthlpy change of combustion thus it will be exothermic.so A is wrong and we can never judge a reaction type by its magnitude differance,a reaction can be +1000 or -1000 thus sign matters.so alone 1000 cant give us the detail.

How to solve these MCQ's, June 2003, MCQ 2, 8, 10, 12, 18, 27, 33, 29.
In mcq 32 options 1 and 3 r correct so obviously the 2nd one is correct as well, but if we r not sure about the 1st and 3rd options so how do we know that the alkyl chain is soluble in oil droplets??

 

[!>@/^~^\@<!]

How to solve these MCQ's, June 2003, MCQ 2, 8, 10, 12, 18, 27, 33, 29.
In mcq 32 options 1 and 3 r correct so obviously the 2nd one is correct as well, but if we r not sure about the 1st and 3rd options so how do we know that the alkyl chain is soluble in oil droplets??

Oh somebody else has already posted all the ans of June 2003, i'll shortly post my other problems

 

[itallion stallion]

How to solve these MCQ's, June 2003, MCQ 2, 8, 10, 12, 18, 27, 33, 29.
In mcq 32 options 1 and 3 r correct so obviously the 2nd one is correct as well, but if we r not sure about the 1st and 3rd options so how do we know that the alkyl chain is soluble in oil droplets??

Q2=see the question,it says one of its side chain is changed to the product shown,in the product you will see that there is only two double bond while in reactant there were 3 so one double is finished so it means one mole of H2 would be used,in the same way the two chain have on double bond so they have lost two so we would need 2mole H2 per chain to break 4 double bonds .so total 5 mole required.
Q8:in part a change is +5 as NH3 has -3 and NO will have +2 so +5.in rest we have +1 in partb,+2 in part c,+3 in d.
Q10:H2+I2=2HI
intial(0.20),(0.15),(0)
so 0.26 moles of HI are formed.H2:HI is 1:2 so if o.26 are formed so 0.26/2 will be used for both H2 and I2.moles left will be (0.20-0.13=0.07) for H2,0.02 for I2.write kc equation remember to take square for HI.
Q12:simply learn this fact.
Q18option c is correct 2Ca(OH)2+NH4(SO3)2 gives 2CaSO4+NH3+H2option a will give ammonium bromide, b will give ammonium choloride and d will have no reaction
In Q27 remember that Cl destroys the ozone layer most.so always Cl will be removed,so C.
Q33 dont know.
Q29:in order to get ethene from alcohol we have to reduce it so H2SO4 will reduce the alkene and then we will react it with NaOH to neutralize and acid present.
Q32:always remember that large carbon chain are soluble in oil.and polar parts are soluble in water.

 

[ahmed abdulla]

Q2=see the question,it says one of its side chain is changed to the product shown,in the product you will see that there is only two double bond while in reactant there were 3 so one double is finished so it means one mole of H2 would be used,in the same way the two chain have on double bond so they have lost two so we would need 2mole H2 per chain to break 4 double bonds .so total 5 mole required.
Q8:in part a change is +5 as NH3 has -3 and NO will have +2 so +5.in rest we have +1 in partb,+2 in part c,+3 in d.
Q10:H2+I2=2HI
intial(0.20),(0.15),(0)
so 0.26 moles of HI are formed.H2:HI is 1:2 so if o.26 are formed so 0.26/2 will be used for both H2 and I2.moles left will be (0.20-0.13=0.07) for H2,0.02 for I2.write kc equation remember to take square for HI.
Q12:simply learn this fact.
Q18option c is correct 2Ca(OH)2+NH4(SO3)2 gives 2CaSO4+NH3+H2option a will give ammonium bromide, b will give ammonium choloride and d will have no reaction
In Q27 remember that Cl destroys the ozone layer most.so always Cl will be removed,so C.
Q33 dont know.
Q29:in order to get ethene from alcohol we have to reduce it so H2SO4 will reduce the alkene and then we will react it with NaOH to neutralize and acid present.
Q32:always remember that large carbon chain are soluble in oil.and polar parts are soluble in water.

can u help bro

 

[Lyfroker]

Q3:

n=50/2x(23+3x14) [2x because there are 2 moles NaN3.]
n= 5/13

3n=V/24 [3x because there are 3 moles of N2.]
3x5/13 x 24 =V
27.69 = V

The answer is (C) which is 27.7


Q9:

One molecule of HCl dissipates one H+ ion. Now looking at the options:
(A) Ethanoic Acid [It is weak acid and cannot dissipate the H+ ion completely.]
(B) Nitric Acid [The formula is HN)3. Each molecule can dissipate one H+ ion, same as HCl.]
(C) Sodium Hydroxide [It doesn't even dissipate H+ ion!!.]
(D) Sulphuric Acid [H2SO4 dissipates 2 H+ ions per molecule, twice that of HCl.]

So the answer is (B),


Q18: [I couldn't do it. Will try again soon.]

Q37:

1 All bond angles are approximately 120o. [This is true. I hope you know why.]
2 It will undergo electrophilic addition reactions. [Yes they undergo electrophlic addition because of presence of double bonds.]
3 It will undergo nucleophilic addition reactions. [Yes they undergo nucleophlic addition because of presence of lone pairs on Oxygen.]

So answer is (A).

Q39:

First of all, alkanes are not soluble. Arene rings are alkanes which aren't soluble. This gives direct answer of option 1 being correct. Option 1 is a Aldehyde which are oxidized to Carboxylic acids, which are quite soluble! So answer is (D).
Note: Option 2 is also an aldehyde and form a carboxylic acid but with poor solubility due to presence of arene ring. Option 3 is a Ketone and cannot be easily oxidized further.

(If I'm wrong somewhere, please correct me. I got my answers this way which are all correct.)

Thnx a lot
Actually in q#37 won't the C=C be 180 n also C=O ?

 

[Lyfroker]

q# 3, 9, 10, 23,28 & 32
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf

 

[itallion stallion]

can u help bro

I gave it a try,CH3CH2CH2CH=CH2,CH3CH2CH=CHCH3(it will two cis, and trans),CH2=CHCH(CH3)CH2CH3 thats 3 methyl pent-1-ene.it will have cis trans as well as structural isomerism.
Q29:when we will add cold KmnO4.alkene will form diol.so there willl be increase in chiral carbons by twn addtion of hot KmnO4 one chiral will decrease because the OH on the left hand side will also oxidise to CO2H,while the alkene carbon will also oxisdise to CO2H so won't remain chiral.hope u got it.

 

[itallion stallion]

M/J/11/11 q6,8,13.
Can someone pls explain?

In question 6 the curve y shows that more O2 is produced which means that more H2O2 would have been used.if the both yield would be same then we could say that less catalyst or low temperature was used.if yield was less than definately water was added.Le chatliers rule works here.
Q8X-X)=-2775+6(C-H)=+405.5
Q13:4Al+3O2=2Al2O3 so Al:Al2O3 is 2:1 so moles of aluminium oxide will be 0.01moles.
Al2O3+6HCl=AL2Cl6+3H2O.so ratio is 1:6 SO moles of Hcl will be 0.06 moles.divide it by its concentration and multiply it by thousand u will get the answer

 

[daisyy]

Can anyone explain me question number 20 in http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf thanks!

 

[cute97]

Anyone here to help me with these pls
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q16, Q17 what do they mean by ligand here ?? Q24, Q30 the answer is hydrolysis but its not condensation polymerisation and if it is condensation how did u knw?? and Q32 what is planer here ??
Thank you in advance

 

[1597.2217]

Anyone here to help me with these pls
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q16, Q17 what do they mean by ligand here ?? Q24, Q30 the answer is hydrolysis but its not condensation polymerisation and if it is condensation how did u knw?? and Q32 what is planer here ??
Thank you in advance

Q16:

Gas evolved was CO2 because it turned limewater milky.
CO is present in ALL the options, but remember, that the reactivity of Group II elements increases down the group. Ba is most reactive metals enlisted here, so the answer is (C).

Q17:

I couldn't figure it out myself.

Q24:

No idea.

Q30:

The two monomers are different so it is condensation polymer. (B) is the answer.

Q32:

1- BCl3 is planar.
2- NH3 cannot be due to presence of lone pair on N.
3- PH3 cannot be either due to presence of lone pair on P.

Answer is (D).

 

[cute97]

Q16:

Gas evolved was CO2 because it turned limewater milky.
CO is present in ALL the options, but remember, that the reactivity of Group II elements increases down the group. Ba is most reactive metals enlisted here, so the answer is (C).

Q17:

I couldn't figure it out myself.

Q24:

No idea.

Q30:

The two monomers are different so it is condensation polymer. (B) is the answer.

Q32:

1- BCl3 is planar.
2- NH3 cannot be due to presence of lone pair on N.
3- PH3 cannot be either due to presence of lone pair on P.

Answer is (D).

Thank you and for Q17 someone told me that if you are AS ligands are out of our syllabus

 

[cute97]

Anyone pls
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s05_qp_1.pdf
Q1 and Q2
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q10
thanks in advance

 

[1597.2217]

Well I'm glad that they are out of our syllabus.

Anyone pls
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s05_qp_1.pdf
Q1 and Q2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q10
thanks in advance

ON05.

Q1:

The volume of Oxygen used is Initial-final = 70-20 = 50 Cm^3

This means 10cm^3 of X will react with 50cm^3 of O2 to give 30cm^3 of CO2 and H2O
With this we can make a equation giving the moles:

X + 5 O2 ---> 3 CO2 + y H2O [We can find y by 5x2 O2 - 3x2 O2 giving us 4 O which are in the H2O. So y=4)
X + 5 O2 ---> 3 CO2 + 4 H2O [Now calculate the number of C and H]

This gives us 3 C and 8 H. So answer is C3H8.


Q2:

From equation 1 we see that 2 moles of Sodium Azide gives us 3 moles of Nitrogen gas. So 1 mole gives 1.5 moles of Nitrogen gas.
From equation 2 we see that 10 moles of Sodium give 1 mole of Nitrogen. However, 2 Mole of Sodium Azide gives 2 Mole of Sodium. So 1 Mole of Sodium Azide will give 1 mole of Sodium. You divide the moles in second equation by 10 to get this:

Na + 0.2KNO3→ 0.1K2O + 0.5Na2O + 0.1N2

So your answer is 1.5 moles from first equation plus 0.1 moles from second equation to give 1.6 moles of Nitrogen gas, which is (B).


MJ06

Q10:

Kp= p2NO2/pN2O2
There are three molecules, 1 on left and 2 on right. So the pressure is shared by 3 molecules. The pressure on left should be 1/3 and on right 2/3. Using these pressure in the Kp equation:
2/3 ^2 / 1/3
We get answer of 4/3 which is (C).

Good luck. Please pray for me, although I can help here I failed P1 in my school exams always.

 

[queen of the legend]

1 In which species does the underlined atom have an incomplete outer shell?

A BF3 B CH3

– C F2O D H3O

why is it not CH3 .............marking scheme ans is A ?!
can any1 explain ?

 

[cute97]

Well I'm glad that they are out of our syllabus.



ON05.

Q1:

The volume of Oxygen used is Initial-final = 70-20 = 50 Cm^3

This means 10cm^3 of X will react with 50cm^3 of O2 to give 30cm^3 of CO2 and H2O
With this we can make a equation giving the moles:

X + 5 O2 ---> 3 CO2 + y H2O [We can find y by 5x2 O2 - 3x2 O2 giving us 4 O which are in the H2O. So y=4)
X + 5 O2 ---> 3 CO2 + 4 H2O [Now calculate the number of C and H]

This gives us 3 C and 8 H. So answer is C3H8.


Q2:

From equation 1 we see that 2 moles of Sodium Azide gives us 3 moles of Nitrogen gas. So 1 mole gives 1.5 moles of Nitrogen gas.
From equation 2 we see that 10 moles of Sodium give 1 mole of Nitrogen. However, 2 Mole of Sodium Azide gives 2 Mole of Sodium. So 1 Mole of Sodium Azide will give 1 mole of Sodium. You divide the moles in second equation by 10 to get this:

Na + 0.2KNO3→ 0.1K2O + 0.5Na2O + 0.1N2

So your answer is 1.5 moles from first equation plus 0.1 moles from second equation to give 1.6 moles of Nitrogen gas, which is (B).


MJ06

Q10:

Kp= p2NO2/pN2O2
There are three molecules, 1 on left and 2 on right. So the pressure is shared by 3 molecules. The pressure on left should be 1/3 and on right 2/3. Using these pressure in the Kp equation:
2/3 ^2 / 1/3
We get answer of 4/3 which is (C).

Good luck. Please pray for me, although I can help here I failed P1 in my school exams always.

THANKKKK YOOOOUU you really made it easy yhh sure wont forget you from my prayers and pray for me and best of luck
dnt worry i got 10 of 40 in the skool exam u will do gr8 ISA

 

[1597.2217]

THANKKKK YOOOOUU you really made it easy yhh sure wont forget you from my prayers and pray for me and best of luck
dnt worry i got 10 of 40 in the skool exam u will do gr8 ISA

You too. In sha Allah.

1 In which species does the underlined atom have an incomplete outer shell?

A BF3 B CH3

– C F2O D H3O

why is it not CH3 .............marking scheme ans is A ?!

can any1 explain ?

Please underline the atom, or provide with the paper number and question number.