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

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i'll answer them again :
q2 : there should be a distribution of energies starting frm zero onwards ...and an increse in temp shifts the molecular energis to the right but area under the two curves should be same since ther no. of molecules remains the same

q:4
use equation of bonds broken - bonds formed .....draw a displayed formula to get the picture clearly

q6: activation energy is between E1 and E2 and since E1 is greater you sbtract the two to get the humped area hieght (the transition state)

q7: this one you have to use ( answer by 6Astarstudent) Here is a quick equation
(V1P1 + V2P2)/(V1+V2) = new P
so (5x12 + 10x6)/(5+10) = 120/15 = 8

q8: draw an enthalpy cycle for easy calculation ....take the vlaue for Ionisation energies of Ca from the booklet

q10:eek:nly graphit is ringed structure here and does not have the tetrahedral structure of the other

q11: write intial moles and equlibrium moles ....multiple the 0.25 of y by 2 since mole ratio is 1:2
to calculate kc (0.25 x 2 )2 / 0.25 = 1
 
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t
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.

Q2 dont know!
Q5.the question says hydrogen bond breaking,h.bonding is only present in ammonia so definately C is the ansewer.
Q10: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
q11:dont know!
Q15:at that temp it will have coordinate bonding in its structure so A is correct structure.
Q17:Cl being more reactive than I will displace it so I will get oxidised.write its equation and u will see.
Q18.Cao being basic nature will stop So2 emmision by forming CaSo4.Learn this use.Dont know regarding A and D.stick to the basics!
Q19 cant explain.
Q28:Hydrazine only reacts with aldehydes and ketones.so D is the answer.
Q35:dont know!!
Q38:eek:ption 1 is cycloalkene so it can take place the added group is at write place,exactly where the double is broken.option 2 is an alkane so this reaction can take place.option 3 is ketone so reaction cant take place.
pray for my result!!!
 
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Q2. always remember that at a higher temperature the curve with the higher temperature will be more to the right of the curve of the lower temperature and will have a lower hump than the curve with the lower temperature so it is B

Q4. again bond breaking is positive and bond forming is negative so the bond which r broken r 1(1077) + 2(436) = +1949 and the bonds which r formed r 3(C-H) = -1230 AND 1(C-o) = -360 and 1(0-H) = -460 so the total bond formed = -1230 -360 - 460 = -2050. enthalpy change of reaction = 1949 + -2050 = -101 so the answer is B
 
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i'll answer them again :
q2 : there should be a distribution of energies starting frm zero onwards ...and an increse in temp shifts the molecular energis to the right but area under the two curves should be same since ther no. of molecules remains the same

q:4
use equation of bonds broken - bonds formed .....draw a displayed formula to get the picture clearly

q6: activation energy is between E1 and E2 and since E1 is greater you sbtract the two to get the humped area hieght (the transition state)

q7: this one you have to use ( answer by 6Astarstudent) Here is a quick equation
(V1P1 + V2P2)/(V1+V2) = new P
so (5x12 + 10x6)/(5+10) = 120/15 = 8

q8: draw an enthalpy cycle for easy calculation ....take the vlaue for Ionisation energies of Ca from the booklet

q10:eek:nly graphit is ringed structure here and does not have the tetrahedral structure of the other

q11: write intial moles and equlibrium moles ....multiple the 0.25 of y by 2 since mole ratio is 1:2
to calculate kc (0.25 x 2 )2 / 0.25 = 1

thank u so much :) this is the bst place for finding awnsers :X3:
 
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Q2. always remember that at a higher temperature the curve with the higher temperature will be more to the right of the curve of the lower temperature and will have a lower hump than the curve with the lower temperature so it is B

Q4. again bond breaking is positive and bond forming is negative so the bond which r broken r 1(1077) + 2(436) = +1949 and the bonds which r formed r 3(C-H) = -1230 AND 1(C-o) = -360 and 1(0-H) = -460 so the total bond formed = -1230 -360 - 460 = -2050. enthalpy change of reaction = 1949 + -2050 = -101 so the answer is B

thankssssss
 
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Which element has an equal number of electron pairs and of unpaired electrons within orbitals of
principal quantum number 2?
A beryllium
B carbon
C nitrogen
D oxygen

Oxygen has 2 paired electrons in its outer shell, one in an s orbital and one in the p. It has 2 unpaired electrons, both in p orbitals.
 
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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 :)
Ok. Thaks alot.:)
 
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19. It's something to be memorised, not deduced.
27. Warmed with sulphuric acid -> ester turns into carboxylic acid and alcohol. In the second step, H2 is added to to all the C=C bonds.
35. You gotta memorise all this. It's just plain facts.
What's the question?

w 10 V12 no. 19c /27a / 35a ---------- done thank u :) 27 didn't realise it was an ester ... i hate skeletal formulas :)

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

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

oculd u please help with the s11 and n11 questions thanks
 
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how many alcohols (including both strructural and stereoisomers) can have the molecular formula c4h10O
 
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HEEEELPPP =3
15 Use of the Data Booklet is relevant to this question.
The combustion of fossil fuels is a major source of increasing atmospheric carbon dioxide, with a
consequential rise in global warming. Another significant contribution to carbon dioxide levels
comes from the thermal decomposition of limestone, in the manufacture of cement and of lime for
agricultural purposes.
Cement works roast 1000 million tonnes of limestone per year and a further 200 million tonnes is
roasted in kilns to make lime.
What is the total annual mass output of carbon dioxide (in million tonnes) from these two
processes?
A 440 B 527 C 660 D 880
 
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HEEEELPPP =3
15 Use of the Data Booklet is relevant to this question.
The combustion of fossil fuels is a major source of increasing atmospheric carbon dioxide, with a
consequential rise in global warming. Another significant contribution to carbon dioxide levels
comes from the thermal decomposition of limestone, in the manufacture of cement and of lime for
agricultural purposes.
Cement works roast 1000 million tonnes of limestone per year and a further 200 million tonnes is
roasted in kilns to make lime.
What is the total annual mass output of carbon dioxide (in million tonnes) from these two
processes?
A 440 B 527 C 660 D 880

write down the equation for decomposition of CaCO3 ---------> CaO + CO2
Mr: 100 g gives 44 g of gas
hence a total of 1200 g limestone will yield 44 x 1200 / 100 = 528 approximatyely B
 
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