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

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf q2,3,4,6,11 actuallly most of the questions of the ppr :p bcz this ppr is kinda hard for me :(

q2: phosphorus is well know ...it has 3 unpaired electrons in its p sub shell : 1s2 2s2 2p6 3s2 3p3

q3: write down the electronic configuration of the three elements

q4:
when bonds are broken energy is absorbed ( + sign) ...when bonds are formed energy is released (- sign)

q6 :
in NH4NO3 ...consider NH4+ .... +4 (hydrogen) + N = +1
so N= -3 in NO3- ... 3* -2 + N = -1 so N= +5
in N2O each N has a ox.state of +1
now calculate change from one nitrogen from NH4 to N in N2O = -3-1 = -4
and change from NO3 to another N in N2O is = 5-1 = +4
hence D :)

q 11o_O Haber process is well known as well ....as pressure increases yield should increase since forward reaction has fewer no. of molecules ...and since the reaction is exothermic increasing the temperature favors backward reaction decreasing the yield of ammonia
 
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i took phys last year but u r lucky that i take mechanics and i still remember this stuff :p anyway the answer is obviously A because the graph A shows the acceleration is constant and when the speed of the car is increasing uniformly it means that the acceleration of the car is constant and doesnt increase. graph B shows that the acceleration is increasing by time so the car will speed up non-uniformly . graph C shows that the car is moving with constant speed ( zero acceleration) and for graph D the car is not moving at all

ANSWER: A
 
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i
i took phys last year but u r lucky that i take mechanics and i still remember this stuff :p anyway the answer is obviously A because the graph A shows the acceleration is constant and when the speed of the car is increasing uniformly it means that the acceleration of the car is constant and doesnt increase. graph B shows that the acceleration is increasing by time so the car will speed up non-uniformly . graph C shows that the car is moving with constant speed ( zero acceleration) and for graph D the car is not moving at all

ANSWER: A
am asking for QUESTION TEN.
 
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relative speed of approach = relative speed of seperation, lots of students get confused by this question because there is no solid formula
note its speed not velocity.
speed of approach for this question you add U1 and U2 because they're coming towards each other.
seperation because they're going in same direction, you use the faster one (in this case V2) minus V1 so correct answer is D

In short, if the masses are in same direction (either before or after collision) you subtract the smaller one from the bigger one.
and if masses are in opposite directions you simply add the speed of the masses
 
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ARE YOU GUYS MAD OR WHAT? I AM ASKING FOR QUESTION TEN. WHY AREN'T YOU GUYS HELPING ME? YOU SELFISH PUMPKINS -_____-
 
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relative speed of approach = relative speed of seperation, lots of students get confused by this question because there is no solid formula
note its speed not velocity.
speed of approach for this question you add U1 and U2 because they're coming towards each other.
seperation because they're going in same direction, you use the faster one (in this case V2) minus V1 so correct answer is D

In short, if the masses are in same direction (either before or after collision) you subtract the smaller one from the bigger one.
and if masses are in opposite directions you simply add the speed of the masses

THANKS A BUNCH MAN!!! I GOT IT ...PLEASE HELP ME IN THIS TOO : http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf QUESTION 7
 
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13 In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.
mHIO → nI2 + pHIO3 + qH2O
This equation can be balanced using oxidation numbers.
What are the values for n and p?

ans : n= 2 and p=1 how do balance using ox . numbers ?
 
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13 In aqueous solution, the acid HIO disproportionates according to the following equation where m,
n, p and q are simple whole numbers in their lowest ratios.
mHIO → nI2 + pHIO3 + qH2O
This equation can be balanced using oxidation numbers.
What are the values for n and p?

ans : n= 2 and p=1 how do balance using ox . numbers ?
ok this is a disproportion reaction
oxidation number of I in HIO is +1, in I2 is 0 and in HIO3 is +5
so HIO to I2 is (+1 -> 0)x2 which is -2 because there is 2 Is in I2.
and HIO to HIO3 is (+1->+5) which is +4
ratio is -2:4 is 1:2 so mol ratio must be 2:1
 
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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.
Q8:(X-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
How to solve these, Nov 2005, MCQ's 2, 7, in mcq 12 shouldn't it be B, 16, 17, 18, 24, 32, 33, 34 35.
Thanks in advance!
 
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ok this is a disproportion reaction
oxidation number of I in HIO is +1, in I2 is 0 and in HIO3 is +5
so HIO to I2 is (+1 -> 0)x2 which is -2 because there is 2 Is in I2.
and HIO to HIO3 is (+1->+5) which is +4
ratio is -2:4 is 1:2 so mol ratio must be 2:1

How to solve these, Nov 2005, MCQ's 2, 7, in mcq 12 shouldn't it be B, 16, 17, 18, 24, 32, 33, 34 35.
Thanks in advance!
 
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question 11, uk the haber process is an exothermic reaction so an increase in the temperature will shift the equilibrium position to the left and the yield of ammonia will be decreased

N2 + 3H2 --> 2NH3 + heat

u have less moles of products than reactants so an increase in the pressure will shift the equilibrium position to the right and therefore the yield of ammonia increases. now use that info to find the answer!

satisfied with my help or there is somethin u dont get still?


thank u so much for ur help :) i wanna ask so many of queshtions from other papers still is that okey :oops:
 
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question 11, uk the haber process is an exothermic reaction so an increase in the temperature will shift the equilibrium position to the left and the yield of ammonia will be decreased

N2 + 3H2 --> 2NH3 + heat

u have less moles of products than reactants so an increase in the pressure will shift the equilibrium position to the right and therefore the yield of ammonia increases. now use that info to find the answer!

satisfied with my help or there is somethin u dont get still?

Can you please ans these, Nov 2005, MCQ's 2, 7, in mcq 12 shouldn't it be B, 16, 17, 18, 24, 32, 33, 34 35.
Thanks in advance!
 
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How to solve these, Nov 2005, MCQ's 2, 7, in mcq 12 shouldn't it be B, 16, 17, 18, 24, 32, 33, 34 35.
Thanks in advance!

Q2:see the OH negative ion.we have electron=8+2=1oe,proton =8+1=9p,and neutron =8+0=8n.so it is the answer.
Q7.the part in which enthalpy change is low it means that someenergy is used to ionize the acid/base.so the enthalpy change of reaction invovving p is low which means it must an weak acid,so ethanoic is the answer,do for rest and u will get the answer.
Q12:Mg+0.5O2=MgO
Al+0.75O2=0.5Al2O3
S8+12O2=8So3.so at first increases but in the reaction wont remain uniform.
Q16:BaCo3 doesnot decompose easily while CaCo3 will decompose to give CO2.so it is the answer.
Q17:dont know!simply learn this.
Q18:ammonia will be given off.dont know what to explain in it.
Q24:dont know.
Q32.nitrogen has a lone pair so it will be pyramidical with 107 angle.so it means will be answer.
Q33:eek:ption 2&3 are common properties of graphite,simply learn the 1st one two.Q
Q34:dont know.
Q35:MgO being ionic and basic and character will have high mp,low thermal conductivity and wont react with basic compounds.hope u got all these.
 
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Can you please ans these, Nov 2005, MCQ's 2, 7, in mcq 12 shouldn't it be B, 16, 17, 18, 24, 32, 33, 34 35.
Thanks in advance!
Q2. D^-1 has 1 proton and 2 electrons(it has a negative charge so it has an extra electron 1+1=2) and 2-1 = 1 neutron so it can't be A
H3O^+ has 3+8 = 11 protons and 3+7 (note there is a positive charge on the oxygen so instead of 8 electrons it will have 7) = 10 electrons and 8 neutrons. so it can't be B either
OD^- has 9 protons and 10 electrons and 9 neutrons. so it can't be C
lets look at the last option OH^- it has 9 protons, 10 electrons and 8 neutrons. we got more electrons than protons and more protons than neutrons so our answer is D

Q7. the stronger the acid/base the higher the enthalpy change will be..so P should be a weaker acid than HCl which is of course will be ethanoic acid so it is either option A/B. Q should be a weaker alkaline than sodium hydroxide because the enthalpy change is lower than the first one, a weaker base will be ammonia. ANSWER: A got it ?
 
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A reaction which causes the presence of oxides of nitrogen in car exhausts is the formation of
NO.
N2 + O2→ 2NO ∆H = +180kJmol–1
What is the bond energy in kJmol–1 of the bond between the atoms in NO?
A 655 B 835 C 1310 D 1670
 
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