chemistry CIE past ppr doubt

[Sue T]

november 2003 (ppr2)
in question 3(bii)they ask us 2 calculate th heat released
n they gave th temp change as 12.2°C
now i used e=mc?t
n  ?t should be in kelvin, so ?t=(12.2 + 273) K
but in the ms, they jus kept it as 12.2 !
why?
& (d) also - i don get it

[nid404]

3b(ii)

In case of water
The specific heat capacity of water is:
o cwater = 1 cal/(g°C) = 1 cal/(g?K)
o cwater = 4.184 J/(g°C) = 4.184 J/(g?K)
o cwater = 4184 J/(kg°C) = 4184 J/(kg?K)

So u shouldn't change the rise in temp to Kelvin...A change in temperature of 12.2
oC is of the same magnitude as a change in temperature of 12.2 K.


For the (d) bit of it
The reaction is Ca+ 2H2O---> Ca(OH)2+ H2
1 mole=40g of Ca give 24dm3 of H2(1 mole of H2...1 mole of any gas at rtp is 24dm3)
therefore 1g of Ca will give 24/40dm3 of hydrogen 0.6dm3 or 600cm3

Hope u've got it

[Sue T]

got it! thank you so much!

[nid404]

got it! thank you so much!

ur welcome

[sweetie]

can any1 help me in solving que.6 paper1 ( AS chem.)?

[nid404]

can any1 help me in solving que.6 paper1 ( AS chem.)?

could u specify which year plz?

[Saladin]

here to help too

[sweetie]

oh, am sooo sorry
its  M/J 2003

[nid404]

PV=nRT is the ideal gas equation
Ideal gases r thosr which have a negligible size and no intermolecular forces
If you need a close-to accurate value of Mr of a real gas....it can tend to ideal when pressure is low..so molecules are further apart and occupy negligible space and when temp is high..so again the molecules are far apart with hardly any interactions(thus satisfying the conditions of an ideal gas and hence the equation)..
So the ans is C

Hope you understood

[sweetie]

thanx alot

cud u also help me in Q3 nov/04  p1??

[vanibharutham]

CH3SH + 3O2 ? CO2 + SO2 + 2H2O
A sample of 10 cm3 of methanethiol was exploded with 60 cm3 of oxygen.

The first task is to find the limiting reagent...

Calculate the number of moles of CH3SH
n = 10/240000 = 0.00041666666667

Calculate number of moles of O2
n = 60/ 24000 = 0.0025
divide by three to get in correct ratio:
0.0008333333

And from this you can see that the oxygen gas was added in excess and so methanethiol was the limiting reagent.

From the equation we know that three moles of oxygen react with one mole of methanethiol...

If we multiply the number of moles of methanethiol by 3, we get 0.00125 moles... this is the number of moles of oxygen that actually reacted...

therefore,
0.00125 moles = 30 cm³

only 30 cm³ of the 60cm³ reacted....

means that 30 was left over...


TOTAL VOLUME OF REMAINING GASES:
10cm³ of carbon dioxide was produced
10cm³ of sulphur dioxide was produced
30cm³ of oxygen was left unreacted...

Hence the answer is
50cm³
 C

[sweetie]

thank u sooo much

[sweetie]

hey ppl,

in da may06 paper1 Q36
why is ammonia not a reducing agent???

and also Q32 of da same year is a bit difficult

thanx in advance  4 any help...

[Saladin]

Question 36

The answer is B. That is both statements 1 and 2 are correct.

1 is true because Cl₂ which has an oxidation state of 0, becomes Cl^-. Thus as reduction is gain of electrons, chlorine is reduced by Ammonia.

2 is true because NH₃ becomes NH₄⁺. It gains a hydrogen ion i.e. a proton. As we know bases are proton acceptors.

Question 32

The answer is B. That is both statements 1 and 2 are correct.

1 is true because you need to isolate an atom of sodium, before you can ionize it, i.e. by removing an electron.

2 is true because, in order for an atom of sodium to gain a charge you need to ionize it. As it has only a charge of 1+, you only take into account the first ionization energy.

[sweetie]

Thank You sooooooo much