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March 29, 2024, 04:25:43 am

Author Topic: VCE Chemistry Question Thread  (Read 2313296 times)  Share 

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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #390 on: February 25, 2014, 07:48:53 am »
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Once you get more into IR and do a lot of problems you will see that the O-H peak is much more broad.  The CH peaks are a lot more defined and sharp (normally jagged as there are way more than one CH in most molecules) these also normally appear bang on 3000. 
--> walk through on IR: http://www.youtube.com/watch?v=uDBWtDASYn4

In infrared spectrosopy, how do you distinquish the different bonds when the peaks in an infrared spectrum are really similar?

For example, I was asked to identify what bond is undergoing a vibrational change in ethanol at 2950cm-1cm
So
The O-H bond occurs at a wavenumber of 2500-3300
The C-H bond occurs at 2850 - 3300

I know both the bonds exist in ethanol and the wave numbers fit for both.. so why is it C-H and also how do they know that it's "streching"?
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Re: VCE Chemistry Question Thread
« Reply #391 on: February 25, 2014, 06:17:58 pm »
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im stuck on what i do for 3c, i know 3d is just 3c x 6.02.

the mol for acetic acid molecules if 0.27. And the answer the says 0.540 moles for 3c, why do they do 0.27 x 2, i know theirs 2 oxygen molecules but still?

Thanks

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Re: VCE Chemistry Question Thread
« Reply #392 on: February 25, 2014, 06:30:17 pm »
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nvm i figured it out haha, stupid question

Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #393 on: February 26, 2014, 04:28:22 pm »
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Could someone please help me with this:

When doing a gravimetric analysis prac, students were asked to collect 1g of Fertiliser and grind it. After grinding it, it was dissolved with water. This solution was filtered and HCl was added to the filtrate. Then, it was heated and after it boiled, BaCl2 was added to form a precipitate BaSO4.

My question is, why was HCl added? Is it because it would cancel out the other ions in the fertiliser so that it wouldn't form precipitates with Barium? or is it because the fertiliser solution was alkaline (since fertilisers contain NH4 and are generally more basic) and so adding an acid would neutralise it?

It would be great if someone helped me out :)


RKTR

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Re: VCE Chemistry Question Thread
« Reply #394 on: February 26, 2014, 04:39:05 pm »
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Adding HCl in gravimetric analysis

googled and found this link
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Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #395 on: February 26, 2014, 07:11:40 pm »
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JadedBlack

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Re: VCE Chemistry Question Thread
« Reply #396 on: February 26, 2014, 07:14:12 pm »
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A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/
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RKTR

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Re: VCE Chemistry Question Thread
« Reply #397 on: February 26, 2014, 08:22:28 pm »
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A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/
mass of Cu=2(63.5)=127 g
mass of Cu/ mass of metal x 100=3.0
mass of Cu = 3.0/100 x mass of metal
127 x 100/3.0 =mass of metal =4233.33 g
% of Cu2S = [2(63.5)+32.0 ]/4233.33 x 100 =3.8%
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JadedBlack

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Re: VCE Chemistry Question Thread
« Reply #398 on: February 26, 2014, 08:30:33 pm »
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thank you, can't believe how simple that was. Shows how much I need some sleep lol
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #399 on: February 26, 2014, 09:07:49 pm »
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A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/

Couldn't you also just say that copper metal itself is 2*63.6/(2*63.6+32) of the copper(I) sulfide present by mass, or 79.9% of the Cu2S, so the % by mass of the copper sulfide must be 3.%/0.799 = 3.75% => 3.8%?
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Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #400 on: February 27, 2014, 06:20:57 pm »
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Hello everyone it'd be great if someone were to help me with this question:

C14H18Cl6N2 is a drug available for the treatment of hypertension. A 1.356g sample of tablets containing this drug was treated to release all the Cl as Cl- ion. The other constituents were filtered off. An excess of silver nitrate solution was then added to the filtrate, forming an AgCl precipitate which was filtered, dried and weighed. Its mass was 1.125 g. Assuming that the drug was the only constituent of the table containing Cl, what was the percentage by mass of the drug in the tablets.

So here is what i am up to:

n(AgCl)= 1.125/143.4 =0.007845 mol since mole ratios are 1:1 I derived that n(Cl)=0.007845 mol

n(Cl) in tablet= 0.007845*6 = 0.04707 mol  [Was this what I was supposed to do? since there were 6 Cl in the drug]

How do I then proceed from here?

Thank You in advance !

RKTR

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Re: VCE Chemistry Question Thread
« Reply #401 on: February 27, 2014, 07:03:41 pm »
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1 mol of the drug gives you 6 mol of Cl-
so u have no. of mol of Cl- ,you divide by 6 to get no.of mol of the drug
then multiply by molar mass to get mass of drug
mass of drug/ 1.356 x100= your answer
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Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #402 on: February 27, 2014, 07:22:40 pm »
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1 mol of the drug gives you 6 mol of Cl-
so u have no. of mol of Cl- ,you divide by 6 to get no.of mol of the drug
then multiply by molar mass to get mass of drug
mass of drug/ 1.356 x100= your answer

Thank you RKTR !  :)

Blondie21

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Re: VCE Chemistry Question Thread
« Reply #403 on: February 27, 2014, 09:07:44 pm »
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Find no of mol of AgNO3
n=cv
  =0.1074(0.01378)
 =0.00148mol
n of Ag+=0.00148mol
n of Cl-in aliquot =0.00148mol
n of Cl- in250ml =250/20 (0.00148)
                         =0.0185mol
n of MCl2 = 1/2 x 0.0185mol
               =0.00925mol

Mass/ no of mol = molar mass
Molar mass= 1.028/0.00925=111.1g/mol
111.1-2(35.5)=40.1= molar mass of the metal ion which is Ca2+

why is this multiplied by a half? I thought it would be 2 because of required/given?
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #404 on: February 27, 2014, 09:22:58 pm »
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Don't rote learn formulas. Think about where they come from.
Pretend magnesium chloride is a box containing two chloride ions. Then, if you had ten chloride ions, you could put them in five boxes. Hence, the number of magnesium chloride units is half the number of chloride ions.
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