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

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these type of questions drive me crazy but i have learnt a formula which will be really helpful in these kind of questions it is Pi = (Ni/Nt ) x Pt

Pi = pressure of the gas
Ni = no of mole of the gas at equilibrium
Nt = total no of moles of both gases at equilibrium
Pt = total pressure

now we have the formula N2O4 --> 2NO2 at the beginning of the reaction before we start we have 1 mole of N2O4 and zero moles of NO2 at equilibrium we will have 0.5 moles because it dissociate by 50% since the ratio is 1:2 we will have 0.5:1 moles so

Pt = given in the question 1 atm
Ni of N204 = 0.5
Ni of NO2 = 1
Nt = 1.5

Pi of N2O4 = 0.5/1.5 x 1 = 1/3 atm
Pi of NO2 = 1/1.5 x 1 = 2/3 atm

Kp = (Pi of NO2)^2/ (Pi of N2O4)
Kp = (2/3)^2/1/3 = 4/3

ANSWER: C

learn the formula it is very helpful with these type of questions
 
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hey can anyone pleaaaaaaasee help me with Question 37 in oct/nov 2006, question 37 in oct/nov 2007 and question 18 may/june 2007!! Thank uu in advance!!
 
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these type of questions drive me crazy but i have learnt a formula which will be really helpful in these kind of questions it is Pi = (Ni/Nt ) x Pt

Pi = pressure of the gas
Ni = no of mole of the gas at equilibrium
Nt = total no of moles of both gases at equilibrium
Pt = total pressure

now we have the formula N2O4 --> 2NO2 at the beginning of the reaction before we start we have 1 mole of N2O4 and zero moles of NO2 at equilibrium we will have 0.5 moles because it dissociate by 50% since the ratio is 1:2 we will have 0.5:1 moles so

Pt = given in the question 1 atm
Ni of N204 = 0.5
Ni of NO2 = 1
Nt = 1.5

Pi of N2O4 = 0.5/1.5 x 1 = 1/3 atm
Pi of NO2 = 1/1.5 x 1 = 2/3 atm

Kp = (Pi of NO2)^2/ (Pi of N2O4)
Kp = (2/3)^2/1/3 = 4/3

ANSWER: C

learn the formula it is very helpful with these type of questions

Thanks alot!!!!!!
 
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foe question 15, first of all study the table and look at each anion reacting with each halogen gas.

lets look at the X^-1 it can reduce both Y2 and Z2 ( very strong reducing agent)
Y^-1 can't reduce none of them so it is obviously a very weak reducing agent
z^-1 can only reduce Y2 and not X2. so from here u can build up ur conclusion..

X^-1 is the strongest reducing agent then comes Z^-1 and the weakest will be Y^-1

ANSWER: B
 
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do esters mix completely with h2o coz in some places to get correct answer this option has to be selected :S
 
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in s10 qp13 question 30 what r the three alkenes that btan2 ol can produce on dehydration i know its's but1-ene and but2-ene what is the third one?
 
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Q2. the equation will be Pb^4+ 2Br^- ----> Pb^2+ Br2
n(pbcl4) = 6.98/349 = 0.02
n(Br2) = 0.02
m(Br2) = 159.8x0.02 = 3.196 g

ANSWER : C

Q9. tricky question here! Ag is solid so it will not be included in our Kc expression at the beginning of the reaction we had 1 mol of both silver ions and iron ions and at equil we have 0.44 mols so 1-0.44 = 0.56 moles of iron(iii) ions r formed

Kc = 0.56/0.44^2
Kc = 2.89 answer is D

question 35: B any halogen under chlorine will react as a reducing agent with H2SO4 so it can't be all three
 
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t
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.
Thanks a lot! How to solve these, June 2006 MCQ's 2, 5, 10, 11, 15, 17, in 18 why not A or D, 19, 28, 35, 38.
 
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hey everybody !! :D

i need some help with these questions please :)

w 10 V12 no. 19c /27a / 35a

s11 V11 no. 7b /10c / 12b /13c ........................ 7 why not A ?

w11 V12 no. 4c /10a / 15d / 17c / 37d ...................... 17 also why not b ?


thank u in advance :) help as much as u can i know they may be alot :)
 
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hey everybody !! :D

i need some help with these questions please :)

w 10 V12 no. 19c /27a / 35a

s11 V11 no. 7b /10c / 12b /13c ........................ 7 why not A ?

w11 V12 no. 4c /10a / 15d / 17c / 37d ...................... 17 also why not b ?


thank u in advance :) help as much as u can i know they may be alot :)

s11v11q 10 MgO has one of the highest melting and boilind point
 
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