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Chemistry and Physics AS paper 12 MCQS

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I meant working!

my friend sender me the working ... but i couldnt get it :p
here it is ... if u get it ..just help me :D
Given that the two gases and amounts, 80%H20, then 20% 2H2 and O2 together in some system at 1 atm, their mole fractions primarily determine the partial pressure. You estimate this by treating the H20 as an ideal gas in that case,

the reaction looks like this:

H20 >> .8H20 + .2H20 >> .8H20 + .2H2 +.1O2

So the total moles is .8 + .2 + .1 = 1.1

H2 mole fraction is .2 / 1.1 = .181818

O2 mole fraction is .1 / 1.1 = .0909

H20 mole fraction is .8 / 1.1 = .7272

Since the total pressure is 1 atm., their partial pressures are analogous to their mole fractions.

H2 partial pressure = .1818 atm

O2 partial pressure = .0909 atm

H20 partial pressure = .7272
 
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May 2007- Q7 is c. looking at the diagram we can see that u + X = v and we know the equation v=u+at hence x is at.
Q37- same resistance because they have the same volume and as the length is same area should be same so overall resistance is also same.
May 2008 - Q21 is A. It was in the syllabus :p just read the derivation from the textbook.
Q28- equation is d*sin theta=n*lambda. d=1/N and n=3 so sin theat/N=3*lambda and sin theta = 3*lambda*N
Q30- its just trignometry in math :p v is hypotenuse, theta is the angle, and u is the horizontal line or the adjacent. cos theta=adjacent/hypotenuse , so v=u/cos theta
May 2011- Q34 is A. resistance is resistivity*area/length. V^1/3 will giv us the length. as it is a cube, area is square of length so area/length will still give length.
May 2012-Q22- mass*6.02*10^23 to give mass of total crystal. then divide by density to give volume. density=mass/volume. then volume/6.02*10^23 to give volume of one atom. then cube root it.
Q24- find the area under the graph. Remember to convert mm to m. Area should be between 7mm and 25N :)
 
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May 2007- Q7 is c. looking at the diagram we can see that u + X = v and we know the equation v=u+at hence x is at.
Q37- same resistance because they have the same volume and as the length is same area should be same so overall resistance is also same.
May 2008 - Q21 is A. It was in the syllabus :p just read the derivation from the textbook.
Q28- equation is d*sin theta=n*lambda. d=1/N and n=3 so sin theat/N=3*lambda and sin theta = 3*lambda*N
Q30- its just trignometry in math :p v is hypotenuse, theta is the angle, and u is the horizontal line or the adjacent. cos theta=adjacent/hypotenuse , so v=u/cos theta
May 2011- Q34 is A. resistance is resistivity*area/length. V^1/3 will giv us the length. as it is a cube, area is square of length so area/length will still give length.
May 2012-Q22- mass*6.02*10^23 to give mass of total crystal. then divide by density to give volume. density=mass/volume. then volume/6.02*10^23 to give volume of one atom. then cube root it.
Q24- find the area under the graph. Remember to convert mm to m. Area should be between 7mm and 25N :)
thanks, i feel dumb now though :p

but for q22 they didnt give us any mass except that for a single atom? :/ are we assuming theres going to be 1 mole in the crystal?

and for q24 that is what i did but the answer came out wrong
 
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for q 36 u know for internal resistance cell we find emf by formula E=v-ir .E=3-i*2 lets find i (curret) I=V/R=3/6 6 as total resistance is 6 so we get current as 0.5 amps we get E=3-0.5*2=2
output power means power dissipated in ouput source that is 4 ohm resistor P=(I^2)R =1W so u get it right .
for q 40
find the charge to mass ratio look bigger charge for a certain mass means that a body will accelerate more quickly right ? so find charge to mass ratio of each of these elements u will get C as the lowest ratio with lowest acceleration.
 
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OK .. thank you .. I got question 36, but can you explain q40 a bit more clearly ? . i mean by showing the mathematical functions .. :unsure: ?
 
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thanks, i feel dumb now though :p

but for q22 they didnt give us any mass except that for a single atom? :/ are we assuming theres going to be 1 mole in the crystal?

and for q24 that is what i did but the answer came out wrong
yeah take it as 1 mole because they gave us the density of the whole crystal and not just for 1 atom :p for Q24 find the area of a small block. That will be 1N*(0.5*10^-3). So area of one small block will be 0.0005. Then count the number of blocks in the given region and multiply it by 0.0005 :)
 
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Messages
189
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OK .. thank you .. I got question 36, but can you explain q40 a bit more clearly ? . i mean by showing the mathematical functions .. :unsure: ?
Q40 - For A 1/1 gives 1 ratio. B is 4/2 giving 2 ratio. C is 7/3 =2.3 and D is 9/4=2.25. C has highest ratio :)
 
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