Phys p12 and Chem p12- 9th and 10th June,2015 Tips!

[asadalam]

Hahah rite
Can you solve Q33? I got 0.8ohms. Idk what am I doing wrong.
View attachment 54518

Diameter half so area quarter.Volume constant so Length multiplied by 4.
New expression : P4l/(1/4)A which means P16/l of original
Hence new is 0.2 x 16=3.2

 

[Aly Emran]

Diameter half so area quarter.Volume constant so Length multiplied by 4.
New expression : P4l/(1/4)A which means P16/l of original
Hence new is 0.2 x 16=3.2

Why am I so dumb?

 

[Aly Emran]

Okay here are all qstn: (i'll ask the rest later)
Thanks in advance for helping

Im poor at estimations... Btw whats steel post?

Yar someone plzz tell me how to know when the collision is elastic or inelastic.... I dunno whether head on collision elastic or not... I know their meaning.... KE is not conserved in inelastic... But they ask qstns like what is the change in KE... I thought head on collisions are prefectly elastic :/.... This momentum is so confusing... I cant even visualise it properly!!!
Upthrust always remains constant or only in this scenario it is constant??
Plzz help me with this too

I didntget this qstn properly... The middle resistor is creating confusion :S

 

[Rishab Dev]

Hey Guys this is a question from Chemsitry Paper 12.m/j/2014. Can you help me solve this question? The answer is B.

 

[Mayukh]

Okay here are all qstn: (i'll ask the rest later)
Thanks in advance for helping

Im poor at estimations... Btw whats steel post?
View attachment 54520
Yar someone plzz tell me how to know when the collision is elastic or inelastic.... I dunno whether head on collision elastic or not... I know their meaning.... KE is not conserved in inelastic... But they ask qstns like what is the change in KE... I thought head on collisions are prefectly elastic :/.... This momentum is so confusing... I cant even visualise it properly!!!View attachment 54522
Upthrust always remains constant or only in this scenario it is constant??View attachment 54523
Plzz help me with this too
View attachment 54524
I didntget this qstn properly... The middle resistor is creating confusion :S
View attachment 54525

I'm a bit clueless about the first one. No idea what a steel post is and can't understand what do they mean by the copper wire. All I'm sure of is that the answer is not C or D. By the way which year is it from?

As for the second one, the sum of total momentum=2m*u-m*u=mu
in A, total momentum= -2m*u/3+m*5u/3=mu, so momentum is conserved
in B, total momentum=-2m*u/6+m*2u/3=mu/3, so momentum is not conserved
in C, total momentum=2m*u/6+m*2u/3=mu, so momentum is conserved
in D, total momentum=(2m+m)*u/3=mu, so momentum is conserved

thus the answer is B, where momentum is not conserved

In the third, one the thing to note is that W and U will not change since the volume of air displaced and the gravitational field strength are constant. So, only V will change. At terminal velocity resultant force is 0, so the length of V must negate the resultant after subtracting U from W, simply put the length of U+V hast to equal the length of W. So the answer is D.

The third one I'm not absolutely sure but the answer should be B. This is because the maximum current in the wire will be the same as that in the thinnest section. And, this is the part I'm uncertain of, the velocity of electrons in a specific metal is constant. ( i think i may have read this somewhere or maybe I thought it up myself. Not sure which. Some one better check)


Fourth. Don't look at the Resistor N yet. So p.d across M is 20-7=13V. Now, the potential drop on N is 4V. The potential between L and M is 13V. So the potential between P and Q is 13-4=9V. Since P.d across parallel circuit is 20V, P.d across P is 20-9=11V(If it seems confusing, look at it like this, the 9V came when the reverse was done, i.e 20-11=9V). So p.d across Q is the remaining 9V. So the answer is C.

as for the last one, since there are 2 wires each wire will take half the weight, so young modulus=stress/strain=((40*9.81)/(pi*(2.5*10^-3)))/((1*10^-3)/10)=1.998*10^11 which approximately 2*10^11 so the answer is C.

 

[Mayukh]

View attachment 54540
Hey Guys this is a question from Chemsitry Paper 12.m/j/2014. Can you help me solve this question? The answer is B.

Cyclopropane has molecular formula C3H6. So, making 1 mole requires 3 moles of C atom and 6 moles of hydrogen atom or 3 moles of hydrogen gas. So we can make an equation,
3*dH(C{graphite})+3*dH(H2)-3*dH(C-C)-6*dH(C-H)=53.5
so, 3*dH(C-C)=3*dH(C{graphite})+3*dH(H2)-6*dH(C-H)-53.5
=>3*dH(C-C)=3*717+3*436-6*410-53.5=945.5
=>dH(C-C)=945.5/3=315KJmol^-1, correct to 3 S.F

 

[kesha27]

are there a level chemistry mcq topicals compiled?

 

[Aly Emran]

I'm a bit clueless about the first one. No idea what a steel post is and can't understand what do they mean by the copper wire. All I'm sure of is that the answer is not C or D. By the way which year is it from?

As for the second one, the sum of total momentum=2m*u-m*u=mu
in A, total momentum= -2m*u/3+m*5u/3=mu, so momentum is conserved
in B, total momentum=-2m*u/6+m*2u/3=mu/3, so momentum is not conserved
in C, total momentum=2m*u/6+m*2u/3=mu, so momentum is conserved
in D, total momentum=(2m+m)*u/3=mu, so momentum is conserved

thus the answer is B, where momentum is not conserved

In the third, one the thing to note is that W and U will not change since the volume of air displaced and the gravitational field strength are constant. So, only V will change. At terminal velocity resultant force is 0, so the length of V must negate the resultant after subtracting U from W, simply put the length of U+V hast to equal the length of W. So the answer is D.

The third one I'm not absolutely sure but the answer should be B. This is because the maximum current in the wire will be the same as that in the thinnest section. And, this is the part I'm uncertain of, the velocity of electrons in a specific metal is constant. ( i think i may have read this somewhere or maybe I thought it up myself. Not sure which. Some one better check)


Fourth. Don't look at the Resistor N yet. So p.d across M is 20-7=13V. Now, the potential drop on N is 4V. The potential between L and M is 13V. So the potential between P and Q is 13-4=9V. Since P.d across parallel circuit is 20V, P.d across P is 20-9=11V(If it seems confusing, look at it like this, the 9V came when the reverse was done, i.e 20-11=9V). So p.d across Q is the remaining 9V. So the answer is C.

as for the last one, since there are 2 wires each wire will take half the weight, so young modulus=stress/strain=((40*9.81)/(pi*(2.5*10^-3)))/((1*10^-3)/10)=1.998*10^11 which approximately 2*10^11 so the answer is C.

Thanks a lot
That really helped...
For current even I forgot the ans

 

[kesha27]

Help with these its from paper november p13 2014

 

[mohhef3]

The following half reactions occur when potassium iodate(V), KIO3, in hydrochloric acid solution oxidises iodine to ICl 2 – . IO3 – + 2Cl – + 6H+ + 4e– → ICl 2 – + 3H2O I2 + 4Cl – → 2ICl 2 – + 2e– What is the ratio of IO3 – to I2 in the balanced chemical equation for the overall reaction? A 1: 1 B 1 : 2 C 1 : 4 D 2 : 1
someone try to solbe this and explain plz.paper 12 nov 2013

 

[Aly Emran]

The following half reactions occur when potassium iodate(V), KIO3, in hydrochloric acid solution oxidises iodine to ICl 2 – . IO3 – + 2Cl – + 6H+ + 4e– → ICl 2 – + 3H2O I2 + 4Cl – → 2ICl 2 – + 2e– What is the ratio of IO3 – to I2 in the balanced chemical equation for the overall reaction? A 1: 1 B 1 : 2 C 1 : 4 D 2 : 1
someone try to solbe this and explain plz.paper 12 nov 2013

Answer is B 1:2. Why? Here:

Multiply second equation with 2 so that both equations has 4e-
=(I2+4Cl- ---> 2ICl2- +2e-)*2
=2I2+8Cl- ---> 4ICl2- +4e-

Compare two equations and cancel the electrons:
2I2+8Cl- ---> 4ICl2- +4e-
IO3 – + 2Cl – + 6H+ + 4e– → ICl 2 – + 3H2O

Full equation:
2I2+10Cl- +IO3– +6H+ ------>5ICl2- +3H2O

So ratio IO3-:I2 = 1:2

 

[Aly Emran]

Help with these its from paper november p13 2014

Btw is ans of Q5-B and Q8-D??
If yea than here is what I did:
Q5:
Delta H + 2(-21)=2(-286)+2(-297)
Delta H=-1124kgmol-1

Q8:
moles of Cr2O7=(13.1/1000)*0.100=0.00131mol
moles of Fe2+=6*0.00131=0.00786mol
Mass of Fe2+/Fe=0.00786*55.8=0.4386g

Percentage=(0.4386/1.00)*100=43.9%

I hope they are correct :/

 

[mohhef3]

Answer is B 1:2. Why? Here:

Multiply second equation with 2 so that both equations has 4e-
=(I2+4Cl- ---> 2ICl2- +2e-)*2
=2I2+8Cl- ---> 4ICl2- +4e-

Compare two equations and cancel the electrons:
2I2+8Cl- ---> 4ICl2- +4e-
IO3 – + 2Cl – + 6H+ + 4e– → ICl 2 – + 3H2O

Full equation:
2I2+10Cl- +IO3– +6H+ ------>5ICl2- +3H2O

So ratio IO3-:I2 = 1:2

u r awsome mate thx alot

 

[Aly Emran]

u r awsome mate thx alot

Glad I was able to help u

 

[Rishab Dev]

This is a question from paper 11 2014 o/n. The answer is B. I am calculating the value to be C 4.18*30*200 ryt???? Please explain

 

[kesha27]

Btw is ans of Q5-B and Q8-D??
If yea than here is what I did:
Q5:
Delta H + 2(-21)=2(-286)+2(-297)
Delta H=-1124kgmol-1

Q8:
moles of Cr2O7=(13.1/1000)*0.100=0.00131mol
moles of Fe2+=6*0.00131=0.00786mol
Mass of Fe2+/Fe=0.00786*55.8=0.4386g

Percentage=(0.4386/1.00)*100=43.9%

I hope they are correct :/

Yeah thanks they are the answers
Thank you much ^^

 

[Aly Emran]

View attachment 54643
This is a question from paper 11 2014 o/n. The answer is B. I am calculating the value to be C 4.18*30*200 ryt???? Please explain

here is the solution:
Yes ur doing it right but u didnt calculate the energy of 1mol of fuel

Energy change of 0.0326mol of fuel= mcT
=200*4.18*(55-25)
=25080J

Change for 1mol of fuel=25080/0.0326= 769325Jmol-1 =769KJmol-1

 

[Youknowwhoo]

anybody got a list / methods to memorise molecule shapes ? You know planar pyramid etc ? And bond angles ..: thanks ؟

 

[Aly Emran]

anybody got a list / methods to memorise molecule shapes ? You know planar pyramid etc ? And bond angles ..: thanks ؟

Dont u have to find it out by making structures urself???
Btw can u tell me the bond angle and bond shape of CH3+?

 

[Youknowwhoo]

Dont u have to find it out by making structures urself???
Btw can u tell me the bond angle and bond shape of CH3+?

well trigonal pyramidal I guess man i keep forgetting this stuff .. Seriously you know a compiled note or something would be helpful ;(((