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June 17, 2024, 02:23:09 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 374482 times)  Share 

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saba.ay

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Re: Chemistry 3/4 2013 Thread
« Reply #15 on: December 10, 2012, 05:56:50 pm »
+1
Our chemistry teacher said that the exam will be very similar to the practice exam they put out, seeing as that's what has happened previously. But I guess no one can really be too sure.

Gonna have to study hard and hope for the best. haha

Funkyy007

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Re: Chemistry 3/4 2013 Thread
« Reply #16 on: December 10, 2012, 07:44:09 pm »
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That's true... We all have to study really hard. Welcome to year 12.

bonappler

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Re: Chemistry 3/4 2013 Thread
« Reply #17 on: December 13, 2012, 11:39:25 am »
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Ethylene glycol is used as an anti-freeze in car radiators. It contains carbon, hydrogen and oxygen. Complete oxidation of a 3.10g sample of ethylene glycol produced 4.40g of carbon dioxide and 2.70g of water.
Q. Determine the mass of oxygen in the sample.

teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #18 on: December 13, 2012, 02:41:52 pm »
+1
HI.
I was a bit confused with this question at first. But yeh.
You can find the M.F of Ethylene glycol on the internet , or you can do it like me , by hand.
sorry my camera was pree dodgy

Hope that helps!
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teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #19 on: December 13, 2012, 02:47:43 pm »
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Q. During the production of wine by fermentation , the glucose C6H1206 in grape juice is converted to ethanol( CH3CH2OH) by the action of enzymes from yeast . Carbon dioxide is also produced.
A particular sample of wine is diluted to produce an ethanol concentration of 5.0 v/v. Given the density of ethanol is 0.784 g/mL -1 , determine the concentration of ethanol in the diluted wine sample in mol/L-.

Thanks in advance.

I dont understand how v/v and density are related.
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Lasercookie

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Re: Chemistry 3/4 2013 Thread
« Reply #20 on: December 13, 2012, 03:21:04 pm »
+2
Ethylene glycol is used as an anti-freeze in car radiators. It contains carbon, hydrogen and oxygen. Complete oxidation of a 3.10g sample of ethylene glycol produced 4.40g of carbon dioxide and 2.70g of water.
Q. Determine the mass of oxygen in the sample.
Another way of doing this question is
1. You can find the moles of CO2, so you can find the mass of carbon in CO2. Since all the carbon in the sample is contained in the CO2 afterwards, this would be the same as the mass of Carbon in the sample.
2. You can find the moles of H2O, so you can find the mass of Hydrogen in H2O. Similar reasoning, mass is neither created or destroyed, the bits of Hydrogen in the sample isn't just going to disappear, we can assume that ALL of it will end up in the H2O. Hence this will be the same as the mass of Hydrogen in the sample.
3. The sample is made up of Carbon, Hydrogen and Oxygen. You know the total mass of the sample. You can subtract the mass of carbon and the mass of hydrogen from the total mass of the sample.

For some calculations:
n(C) = n(CO2)
n(H) = 2 * n(H2O)
n(C) in CO2 is the same as n(C) in Ethylene Glycol. This also means that m(C) is also the same (since mass = moles * molar mass).
n(H) in H2O is the same as n(H) in Ethylene Glycol. This also means that m(H) is also the same.
therefore m(O) = m(Sample) - m(C) - m(H)

Number crunching:
Spoiler






You can find the M.F of Ethylene glycol on the internet , or you can do it like me , by hand.
It's probably best to figure it out by hand, yeah. You're not going to have the internet to look up stuff like that in the exam, so you have to go off whatever is given in the question or given in the databook.

Q. During the production of wine by fermentation , the glucose C6H1206 in grape juice is converted to ethanol( CH3CH2OH) by the action of enzymes from yeast . Carbon dioxide is also produced.
A particular sample of wine is diluted to produce an ethanol concentration of 5.0 v/v. Given the density of ethanol is 0.784 g/mL -1 , determine the concentration of ethanol in the diluted wine sample in mol/L-.

Thanks in advance.

I dont understand how v/v and density are related.




Hmm, I'm guessing v/v is percent volume per volume (some type of unit of concentration). This might be an incorrect assumption.

5.0 v/v would mean 5% of the wine is ethanol. So in 100mL of wine, there would be 5mL of ethanol. In 1000 mL of wine, there would be 50mL of ethanol.

Using the density, we can figure out the mass of the ethanol in 1L of wine


 
We can figure out how many moles of ethanol we have here. You can figure out the molar mass of ethanol from the formula given.



Since we just have 1L of solution, the concentration is going to be the same as the number of moles we have


Hopefully I'm not too rusty and didn't make any mistakes.

teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #21 on: December 13, 2012, 04:53:40 pm »
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Another question!
How does oxidisation of alchols occur to make carboxylic acids work?
e.g how does ethanol when added to Cr2O7 2- /H+ make carboxylic acid?
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Re: Chemistry 3/4 2013 Thread
« Reply #22 on: December 17, 2012, 12:41:42 pm »
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HI,
This is a question from Nelson Chem , that I dont understand how to do?

A 50 mL mixture of CO2 and CO gases was mixed with 20 mL O2 gas and exploded. The resulting
gas mixture was bubbled through concentrated LiOH, and 14 mL of gas remained. If all gases were
measured at the same temperature and pressure, what was the percentage by volume of CO in the
original mixture?

When I looked at the question i was like WTF is this? :)
Thanks!!
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thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #23 on: December 17, 2012, 01:06:11 pm »
+1
Hints:

1) Which gas is acidic? LiOH reacts with acids.
2) Volume is proportional to amount.

For answer:
Spoiler
14 mL of gas remains - this means that, since CO2 would react with LiOH to form LiHCO3, that 14 mL of gas must be O2. Hence, O2 must have been in excess by 14 mL - so only 6 mL of O2 reacted!

Hence, since 2 CO + O2 --> CO2, this means that there must have been 12 mL of CO that has reacted (with the 6 mL of O2).

Hence, percentage by volume of CO is 12/50 = 24%!
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Re: Chemistry 3/4 2013 Thread
« Reply #24 on: December 17, 2012, 01:42:07 pm »
0
The ionic product of water is 3 x 10–15 at 10 °C. Calculate the pH of pure water at 10° C.
Show your reasoning.

What does the ionic compound of water mean?? :)

thanks and sorry for asking dumb qs.
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thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #25 on: December 17, 2012, 01:47:21 pm »
+1
ionic product of water = [H+][OH-] = 3 x 10^-15. always true at 10 degrees Celsius.

Pure water has [H+] = [OH-] as it self ionises in 2H2O <--> H3O+ + OH-, equal amounts of each product.

Can u do it from here?

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Homer

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Re: Chemistry 3/4 2013 Thread
« Reply #26 on: December 18, 2012, 03:37:32 pm »
+1
for a 0.20M solution of potassium sulfate k2so4, calculate the amount in mol of:
a)potassium ions
b) sulfate ions
c) oxygen atoms
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Re: Chemistry 3/4 2013 Thread
« Reply #27 on: December 18, 2012, 03:51:14 pm »
+1
@toucan: thanks for your help!! :)

@ jai: dont you need the volume??
ahh now i realise :
0.2M=0.2mol/L-
Therefore,
K+ = 0.2 x 2(as there are 2 K+ atoms)=0.04
SO4 2-= 0.2 X 1=0.02
oxygen= 0.2 X 4= 0.08


Hope that helps!!
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Homer

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Re: Chemistry 3/4 2013 Thread
« Reply #28 on: December 18, 2012, 03:55:21 pm »
+1
its actually, (0.2/39.1) times by 2 since there are 2 K+ atoms and so forth,  however with the so4 ions i get 0.0021 instead of 0.05mol :/
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teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #29 on: December 18, 2012, 04:10:35 pm »
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 ;D HAHA. my bad. confused my self. oh well, learning experience!

thanks !
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i liek lala :) arre bhaiya aal izz well :) <3