ALL CIE CHEMISTRY DOUBTS HERE !!

[Dania]

Thanks I have a MCQ also:

The density of ice is 1.00 g cm^–3.

What is the volume of steam produced when 1.00 cm³ of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?

[1 mol of a gas occupies 24.0 dm³ at 25 °C (298 K) and one atmosphere.]
A 0.267 dm³
B 1.33 dm³
C 2.67 dm³
D 48.0 dm³

The answer is C. How would I obtain it? I'm not sure what equation to use.

[Deadly_king]

Thanks I have a MCQ also:

The density of ice is 1.00 g cm^–3.

What is the volume of steam produced when 1.00 cm³ of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?

[1 mol of a gas occupies 24.0 dm³ at 25 °C (298 K) and one atmosphere.]
A 0.267 dm³
B 1.33 dm³
C 2.67 dm³
D 48.0 dm³

The answer is C. How would I obtain it? I'm not sure what equation to use.

First you need to find the number of moles present in 1cm³ of ice.

Using density = 1gcm^-3, we take the mass present in 1 cm³ of ice to be 1g.

Number of moles = Mass/Mr ----> Number of moles = 1/18 = 0.0556

Now we use the formula for ideal gas ---> PV = nRT

V = (0.0556 x 8.31 x 596)/(101 x 10³) =  0.0027m³

This is equal to 2.7dm³

Am not sure though.........since I did not obtain 2.67 dm³ 

[nid404]

You've become faster

[Deadly_king]

You've become faster

Hehe......yeah I guess so 

Anyway it would be nice if you could confirm the method since I didn't obtain the exact answer 

[Dania]

The answer is D.
Why?

[Dania]

The answer is B. Why?

[Deadly_king]

The answer is D.
Why?

Since it is unreactive to mild oxidising agents, the compound should have been a tertiary alcohol before dehydration. Only D offers this possibility.

NOTE : Tertiary alcohols cannot be oxidised.

[ashish]

The answer is B. Why?

CaCO3(limestone)  -----> CaO + CO2

from equation 1 mole CaCO3 produced 1 mole CO2


Mr CaCO3 =100.1

Mr CaO=56.1

Mr CO2=44

1000mt CaCO3 will give 560.1mt Cao and 440mt CO2
200mt  CaCO3 will give 112.02mt Cao and 88mt CO2

mt=million tonnes

total CO2 = 440+88
=528 --- near to 527

IS that CIE i never worked out such question before
i am not sure whether the method is good or not

[Dania]

Ah, yes. I think I didn't read the question properly, I get it now. Yeah, it's a CIE paper question.

[Deadly_king]

CaCO3(limestone)  -----> CaO + CO2

from equation 1 mole CaCO3 produced 1 mole CO2


Mr CaCO3 =100.1

Mr CaO=56.1

Mr CO2=44

1000mt CaCO3 will give 560.1mt Cao and 440mt CO2
200mt  CaCO3 will give 112.02mt Cao and 88mt CO2

mt=million tonnes

total CO2 = 440+88
=528 --- near to 527

IS that CIE i never worked out such question before
i am not sure whether the method is good or not

Good job pal........though it was a bit long 

+ rep

[moon]

plz help in Nov.2009 ppr22 Q2(v) &

also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)

thanx in advanc.

[moon]

can u plz another question?

complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon?

[elemis]

can u plz another question?

complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon?

Your question is incomplete I need the mass of the original hydrocarbon.

But the methodology is as follows :

If there are 2 grams of Hydrogen in 18 grams of water   THEN

there are    X grams of Hydrogen in 0.072 grams of water

Hence, X = 8*10^-3

If there are 12 grams of Carbon in 44 grams of CO₂ THEN

there are     Y grams of Carbon in 0.352 grams of CO₂

Hence, Y = 0.096

At this point you would add X and Y and subtract that value from the mass of the Hydrocarbon to find the mass of oxygen.

Then you would have the mass of C H and O

You could then calculate the Empirical formula.

[Deadly_king]

can u plz another question?

complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon?

Nan Ari......the data provided is sufficient to find the answer.

C_xH_y + (x + y/4)O₂ ----> xCO₂ + (y/2) H₂O

Number of moles CO₂ = Mass / Mr = 0.352/44 = 0.008

Number of moles of H₂O = 0.072/18 = 0.004

1 mole of the hydrocarbon gives 0.008 moles of CO₂ and 0.004 moles of H₂O

Therefore from equation above we can note that x = 0.008 while y/2 = 0.004

Empirical formula will hence be C₈H₈

If you don't understand, let me know

[elemis]

Ah, damn. I didnt think of it in that way. 

+rep.