6 replies to this topic

[Azie]

im stuck at a few calculation questions from various past papers so any help wud b greatly appreciaited


2001 - Q1©
Technetium-100 has a half-life of 16s. If a sample of technetium-100 is left for 48s, what fraction of the sample would remain?
i have no idea how 2 even begin this


2002 - Q10(d)
N2(g) + 3H2(g) ---> 2NH3(g)
Under certain conditions, 200kg of hydrogen reacts with excess nitrogen in the Haber Process to produce 650kg of ammonia. Calculate the percentage yield of ammonia.


2002 - Q12(b)(i)
2NaHCO3(s) ---> Na2C)3(s) + CO2(g) + H2O(g)
Calculate the theoretical volume of carbon dioxide produced by the complete decomposition of 1.68g of sodium hydrogencarbonate. (Take the molar volume of carbon dioxide to be 23 litres mol(-1).


2002 - Q14(b)(ii)
In a second experiment it was found that 1.2 x 10 to the power -5 mol of iodine reacted with 3.0cm3 of the sodium thiosulphate solution. Use thi information to calculate the concentration of the sodium thiosulphate solution, in mol l(-1).

[Glitterandtrauma]

2001 - Q1©
Technetium-100 has a half-life of 16s. If a sample of technetium-100 is left for 48s, what fraction of the sample would remain?

it tells you the mass of technetium is 100, it also tells you your leaving it for 48s that makes it 3 half lifes ( 3 x 16)

Make sure you dont just divide 100 by 3 though it doesnt work like that. The sample halfs, halfs again and then one more so youve got:

100g to 50g to 25g to 12.5g To get your fraction you simply do : 12.5g divided by 100(mass you started with) which gives you 0.125.

Hope thats clear im not so good at explaining hehe

[Azie]

yeah that was clear, an i understand it all now, thanks i jus didnt wanna leave it out incase sumthin like it came up in the exam an i kicked myself 4 not figurin it out be4 even tho it is only worth 1 mark, my teacher always says that 1 mark cud b the difference between an A and a B.

[englisher]

2002 - Q10(d)
N2(g) + 3H2(g) ---> 2NH3(g)
Under certain conditions, 200kg of hydrogen reacts with excess nitrogen in the Haber Process to produce 650kg of ammonia. Calculate the percentage yield of ammonia.

Well number of moles of hydrogen = mass/GFM = 200/2 = 100 thousand (thousand as 1kg = 1000g)

As 3H2s give 2 Ammonias, you can see that 100 / 3 * 2 = 66.666thousand

Therefore you'd expect to get 66.66thousand moles of ammonia. So that would have a mass of 66.666... * (14+1+1+1)=1133thousand g = 1133kg

Therefore percentage yield = actual mass/ theoretical mass = 650 / 1133 = 57.4%

It's easier to ignore the thousands as it will remain thousands all the way through

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Use this as an example to try Q12b yourself! Remember that molar volume = volume / number of moles.

Any problems just ask

[Azie]

thanks a lot 4 Q10, i wud never have thot of doin that ...gonna attempt Q12 now again so i'll let u no how that goes an where i'll need help hehe

[Hassan]

2002 - Q14(b)(ii)
In a second experiment it was found that 1.2 x 10 to the power -5 mol of iodine reacted with 3.0cm3 of the sodium thiosulphate solution. Use thi information to calculate the concentration of the sodium thiosulphate solution, in mol l(-1).

In the equation you have 1 mole of iodine reacting with 2 moles of thiosulphate ions. So you have a ratio of 1:2 which means you have 2.4 x 10 to the power -5 mol of thiosulphate ions.
Then you use the n = c x v equation, but you need to find out the concentration so its c = n / v

--> 2.4 x 10 to the power -5 / 0.003

= 0.008 mol/l

[Azie]

ahh the working seems so simple compared to my initial thoughts of the question ...thanks a lot