Chemistry MCQ thread...

[sniper7137]

@smartangel

Use the equation:

CaCO3 ---> CaO + CO2

Total of 1200 x 10^6 x 10^3 kg of CaCO3 are roasted

So No. of moles = (1200 x 10^6 x 10^3) / (40 + 12 + 48) = 1.2 x 10^10 mol.

CaO : CO2
= 1 : 1

So No. of moles of CO2 produced = 1.2 x 10^10 mol

Therefore mass of CO2 produced = (1.2 x 10^10) x (12 + 32) = 5.28 x 10^11 kg = 528 million tonnes.

SO THE ANSWER IS B.

 

[Hateexams93]

Can some1 help me ?????? my Q was
Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate
nitrogen gas, which inflates the bag.
2NaN3
3N2 + 2Na
A typical driver’s airbag contains 50 g of sodium azide.
Calculate the volume of nitrogen this will produce at room temperature.
A 9.2 dm3 B 13.9 dm3 C 27.7 dm3 D 72.0 dm3

 

[Zishi]

Can some1 help me ?????? my Q was
Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate
nitrogen gas, which inflates the bag.
2NaN3
3N2 + 2Na
A typical driver’s airbag contains 50 g of sodium azide.
Calculate the volume of nitrogen this will produce at room temperature.
A 9.2 dm3 B 13.9 dm3 C 27.7 dm3 D 72.0 dm3

Convert grams of sodium azide into moles, then take ratios.

NaN3 : N2
2 : 3
5/Mr : x

x = moles of N2

Volume of N2 evolved = 24 x moles

 

[Hateexams93]

i did and i got the wrong answer :/, can u plz do it ?

 

[Zishi]

YA its b ...i also got it...bt was a bit tricky

Post more and more tough question, please!!

YA its b ...i also got it...bt was a bit tricky

how did u get it ?

I'm waiting for hassam to explain it to you. I want to see whether he did it by my method OR any other one. :roll:

 

[Zishi]

i did and i got the wrong answer :/, can u plz do it ?

I get C as the answer. Is that correct?

 

[Hateexams93]

yes , but how ?

 

[Zishi]

yes , but how ?

Okay, so moles of NaN3 = 50/(23+(14x3)) = 10/13

Moles of N2 = (10/13) times (1.5) = 15/13 moles of N2

Volume of N2 = 15/13 times 24 = 27.7 dm^3

 

[Xthegreat]

YA its b ...i also got it...bt was a bit tricky

Post more and more tough question, please!!

YA its b ...i also got it...bt was a bit tricky

how did u get it ?

I'm waiting for hassam to explain it to you. I want to see whether he did it by my method OR any other one. :roll:

try mine. tell me if you guys used the same method

 

[Hateexams93]

16 Chlorine compounds show oxidation states ranging from –1 to +7.
What are the reagent(s) and conditions necessary for the oxidation of elemental chlorine into a
compound containing chlorine in the +5 oxidation state?
A AgNO3(aq) followed by NH3(aq) at room temperature
B concentrated H2SO4 at room temperature
C cold dilute NaOH(aq)
D hot concentrated NaOH(aq)

 

[Xthegreat]

D hot concentrated NaOH

3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O

Cl in NaClO3 is +5

- in A and B is -1
- C is -1 and +1

 

[xHazeMx]

i cant solve this question :/ . can anyone explain it?

 

[Zishi]

YA its b ...i also got it...bt was a bit tricky

Post more and more tough question, please!!

try mine. tell me if you guys used the same method

Yep. I used the same.

 

[Zishi]

i cant solve this question :/ . can anyone explain it?

See "Xthegreat"'s image. He's done it in the same way as I did.

 

[xHazeMx]

i dont understand how did he get 1.089 x 10^9 , can u explain this question zishi ?

 

[Zishi]

i dont understand how did he get 1.089 x 10^9 , can u explain this question zishi ?

Its because we've got an equilibrium constant for [O]^(1/2) but we want to change it into [O}^1 so we take its square, and so also by taking square of 3.3x10^4 we get 1.089x10^9

 

[xHazeMx]

yea right ! .. thanks

 

[xHazeMx]

how about the substitution part ( 6.2/1.089 x 10^9 ) ???

 

[xHazeMx]

Give this a try :%)

 

[Zishi]

Give this a try :%)

I'd say that D is the answer.