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Author Topic: Water's Noobie Chem Thread.  (Read 10013 times)  Share 

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Water

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Re: Water's Noobie Chem Thread.
« Reply #30 on: April 12, 2011, 01:05:41 am »
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Give Brief Explanation for each of the following:

b) When a stick of chalk is placed upright into a beaker that contains a small quantity of black ink, various colours are observed to move up the chalk

My Explaination: The stick of chalk acts as a stationary phase, where the black ink will act as both moving phase and solvent. The Different and adsorption rates will separate the components within the black ink.


Is this appropriate for an answer, or is it flawed?.
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When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

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Re: Water's Noobie Chem Thread.
« Reply #31 on: April 12, 2011, 11:10:28 am »
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Quote from: Water on April 12, 2011, 01:05:41 am
The Different and adsorption rates will separate the components within the black ink.
Hmm your answer seems ok, but maybe include a bit more detail. Is this an exam quesiton or textbook question?
Perhaps change that into something like: the components of the ink continually undergo adsorption onto the stationary phase followed by desorption back into the mobile phase. The degree to which this occurs causes the different components of the ink to separate.
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Water

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Re: Water's Noobie Chem Thread.
« Reply #32 on: April 12, 2011, 04:07:32 pm »
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Just a textbook question xD
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When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

Water

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Re: Water's Noobie Chem Thread.
« Reply #33 on: April 21, 2011, 08:19:30 pm »
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When 1.00g of a mixture containing the minerals As4S6 and As4S4 was completely burnt in excess oxygen, the products formed were As4O6 (s) and SO2(g). The solid product remaining had a mass of 0.905g.

a) Write balanced chemical equations for the reaction of both As4S6 and As4S4 with oxygen.

b) Calculate the mass of both As4S6 and As4S4 in the original Mixture.


So for my answers, I had m(As4S6) = 0.531g    and for m(As4S4) = 0.469g.... 


I would like to consult with VN, to see if I"m correct. Thankyou :)
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About Philosophy

When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

luken93

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Re: Water's Noobie Chem Thread.
« Reply #34 on: April 21, 2011, 09:00:35 pm »
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As4S6(aq) + As4S4(aq) + 16O2(g) --> 2As4O6(s) + 10SO2(g)[/tex]

(The solid is Arsenic Hexoxide)

b)









Hmm, still getting over a gram. Whether they aren't using VCAA molar masses, I'm not really sure...
« Last Edit: April 21, 2011, 09:41:07 pm by luken93 »
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Water

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Re: Water's Noobie Chem Thread.
« Reply #35 on: April 21, 2011, 09:08:00 pm »
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Luken, I did your method, however it yielded greater than 1gram. Hence, that method couldn't be used xD....

My method was abit tedious, wasn't sure, if how it was set out was right. I generally lose marks for setting out, cause my teacher is a dick like that.

n(As4O6) = 0.905g / 395.68gmol
               = 0.002287

m(As) in (As4O6) = 0.002287 x 4 x 74.92gmol-1
                        =0.6854g


m(S) in mixture = 1.00g - 0.6854g
                     = 0.3146g



m(As) in (As4S6) = 1/2 x 0.6854g
                        = 0.3427g

 
m(S) in As4S6 = 6/10 x 0.3146g
                   = 0.18876gram

m(As4S6) = 0.3427g + 0.18876g
              = 0.531gram


m(As) in (As4S4) = 1/2 x 0.6854g
                        = 0.3427gram


m(S) in As4S4 = 4/10 x 0.3146g
                    = 0.1258gram

m(As4S4) = 0.3427g + 0.1258gram
             = 0.469gram



m(As4S4) + m(as4S6) = 0.531g + 0.469gram = 1.00gram
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About Philosophy

When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

luken93

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Re: Water's Noobie Chem Thread.
« Reply #36 on: April 21, 2011, 09:24:38 pm »
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Woops, didn't even realise it had to add up to one gram =/ But as I said, perfect rounding/sig figs would've got the correct answer...

Although, couldn't you just work out the ratios of the molar masses of the two?

Mr(As4S6) = 492.2
Mr(As4S4) = 428.0

Ratio As4S6 : As4S4
 = 1.15 : 1
 = 0.53488 : 0.46511

In the end, that is what you are trying to find out, the mole ratios are the same?

 
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Water

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Re: Water's Noobie Chem Thread.
« Reply #37 on: April 21, 2011, 09:29:07 pm »
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Sorry for my noob question

Ratio As4S6 : As4S4
 = 1.15 : 1
 = 0.53488 : 0.46511


How'd you get from 1.15, to 0.53... and 1 to 0.46511, mathematically


I Knew there was a much more efficient way than mines. Mine was just pathetic if it was examination period Lol.


And, would mine be considered wrong? Cause answers look different to yours, and I used more calculations, hence more room for error.
« Last Edit: April 21, 2011, 09:36:13 pm by Water »
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About Philosophy

When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

luken93

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Re: Water's Noobie Chem Thread.
« Reply #38 on: April 21, 2011, 09:36:59 pm »
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First question: I multiplied 1 x (1.15/2.15) and 1 x (1/2.15), or alternatively you could have multiplied it by the sum of the Mr's multiplied by the Mr of the one you are trying to find, just like when you are finding the mass of hydrogen in water, 1 x (2/18)...

Ummm, I'm not really sure who's wrong. I think it may be best if I head back and do mine exactly again. Stay posted
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Water

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Re: Water's Noobie Chem Thread.
« Reply #39 on: April 24, 2011, 04:20:38 pm »
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Hey guys, do we have to remember the structure of glucose, fructose and galactose >_<?
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About Philosophy

When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

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Re: Water's Noobie Chem Thread.
« Reply #40 on: April 24, 2011, 04:36:26 pm »
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I dunno but the strucutres for all the amino acids and certain biomolecules such as sucrose, glycerol, deoxyribose, adeninel guanine, cytosine, thymine and phosphate are in the data booklet!!
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luken93

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Re: Water's Noobie Chem Thread.
« Reply #41 on: April 24, 2011, 06:14:25 pm »
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If I can remember correctly, fructose has 2x CH2OH groups, and then differentiating glucose and galactose involves looking at the position of the OH groups on the left hand side...
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Water

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Re: Water's Noobie Chem Thread.
« Reply #42 on: May 19, 2011, 09:31:52 pm »
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A soluble fertiliser contains phosphorus in the form of phosphate ions (PO4 3-). To determine the PO4 3- content by gravimetric analysis, 5.97 g of fertiliser powder was completely dissolved in water to make a volume of 250.0ml. A 20.00 volume of this solution was pipetted into a conical flask and the PO4 3- ions in the solution were precipitated as MgNH4PO4. The precipitate was filtered, washed and then converted by heating into Mg2P2O7. The mass of Mg2P2O7 was 0.0352g.


Calculate the amount, in mole, of phosphorus in the 20.00ml volume of solution.


So,


My method of calculation is, find mol of Mg2P2O7.


Then n(MgNH4PO4) = n(Mg2P2O7)

                           
 n (P) = n (MgNH4PO4) 


Am I correct in my method?
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Water

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Re: Water's Noobie Chem Thread.
« Reply #43 on: May 19, 2011, 10:42:31 pm »
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Bump*
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When I see a youth thus engaged,葉he study appears to me to be in character, and becoming a man of liberal education, and him who neglects philosophy I regard as an inferior man, who will never aspire to anything great or noble. But if I see him continuing the study in later life, and not leaving off, I should like to beat him - Callicle

Mao

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Re: Water's Noobie Chem Thread.
« Reply #44 on: May 20, 2011, 12:18:58 am »
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Not reading the question, but I would have thought

n(MgNH4PO4) = 2 n(Mg2P2O7) ?

Anyways, a more direct method would be

n(P) = 2 n(Mg2P2O7)
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