ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: HungTran2009 on June 05, 2010, 08:57:05 pm
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CSE 2010
A) 60 base pairs of which 30 are guanine
B) 50 base pairs of which 30 are cytosine
C) 50 base pairs of which 30 are adenine
D) 45 base pairs of which 20 are thymine
The answer is A.
At first glance I thought about G and C because of the 3 hydrogen bonds that are formed between them. But why would A be the answer over B? Is it because 60 is greater than 50...or??!!
Thanks
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Yeah 60 > 50, more bonds to break after taking into consideration CG ones.
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Hurrah! Thank you :D
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I love these style of questions, easy - quick :)
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very nice question, I though B due to percentage forgetting about total number of bond. I was expecting to see a trick with them saying a small number of A or T and having people forget that the leftover bonds are in a greater percentage of C and G than the other options provided.
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very nice question, I though B due to percentage forgetting about total number of bond. I was expecting to see a trick with them saying a small number of A or T and having people forget that the leftover bonds are in a greater percentage of C and G than the other options provided.
yeah, i fell for it too.
Does anyone know how in short answer Q4 of this exam, you can distinguish between ethyl ethanoate and methyl propanoate? I can't tell the difference from the proton NMR.
Guessed the right on though haha...
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@ stonecold, I haven't done it yet but I am guessing that you are to look at the actual shift numbers, the peaks that are closer to the Oxygen will have a higher shift than they normally do and the numbers in the data book.
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@ stonecold:
I guess you just need to look at the peak at 2.0 ppm. Refer to the data booklet and you'll see that this is for a methyl group attached to an ester group. Hence ethanoate?
That's just my logic behind it...
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@ stonecold
How did you do question 4 of Multiple Choice??? So...confused :S
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Thanks. I like your logic. I'll type up Q4 now for you Hung. :)
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Thank you! :D
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Okay, so Cr is undergoing reduction right, as it goes from Cr5+ to Cr3+
Fe must therefore undergo oxidation. And it can only go from +2 to +3 state, so initially it must be Iron (II) thus option C is eliminated.
Then find mol Cr2O7 = cv = 0.100 x 0.01278 = 0.001278 mol
To find mol of Fe, you do the half equation:
Fe2+ ---> Fe3+ + e-
Then you multiply by 6, so the electrons cancel out in the full equation.
Ratio Cr2O72- : Fe2+ is 1:6
Therefore, multiply mol by 6.
n(Fe2+)= 0.001278 x 6 = 0.007668 mol
m(Fe2+)= 0.007668 x 55.9 = 0.43 g
Now you test out all the alternatives.
A - Mol Cl is twice the mol of Fe, so 0.007668 x 2 x 35.5 = 0.54
0.54 + 0.43 = 0.97 g - wrong
B - Mol SO4 is the same as the mol of Fe, so 0.007668 x (32.1 +64) = 0.74
0.74 + 0.43 = 1.17 g - wrong
C - we eliminated before
D - Mol NO3 is twice the mol of Fe, so 0.007668 x 2 x (14 + 48) = 0.95
0.95 + 0.43 = 1.38 g - RIGHT!
Absolute pain in the ass question for only one mark...
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WOW :O
Thanks!
And yeah...that's a lot of work for one mark :/
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I got this one straight away, but not sure if my reasoning was correct.
I figured that guanine is a purine, and therefore has a larger molecular mass than none-purines... so B and D are out.
Adenine is generally smaller in molecular mass, but that's irrelevant, 50 is much less than 60 haha
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lol I don't see anything wrong with ur reasoning.
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Okay, so Cr is undergoing reduction right, as it goes from Cr5+ to Cr3+
Fe must therefore undergo oxidation. And it can only go from +2 to +3 state, so initially it must be Iron (II) thus option C is eliminated.
Then find mol Cr2O7 = cv = 0.100 x 0.01278 = 0.001278 mol
To find mol of Fe, you do the half equation:
Fe2+ ---> Fe3+ + e-
Then you multiply by 6, so the electrons cancel out in the full equation.
Ratio Cr2O72- : Fe2+ is 1:6
Therefore, multiply mol by 6.
n(Fe2+)= 0.001278 x 6 = 0.007668 mol
m(Fe2+)= 0.007668 x 55.9 = 0.43 g
Now you test out all the alternatives.
A - Mol Cl is twice the mol of Fe, so 0.007668 x 2 x 35.5 = 0.54
0.54 + 0.43 = 0.97 g - wrong
B - Mol SO4 is the same as the mol of Fe, so 0.007668 x (32.1 +64) = 0.74
0.74 + 0.43 = 1.17 g - wrong
C - we eliminated before
D - Mol NO3 is twice the mol of Fe, so 0.007668 x 2 x (14 + 48) = 0.95
0.95 + 0.43 = 1.38 g - RIGHT!
Absolute pain in the ass question for only one mark...
I agree I did it stonecold's way, up until the last bit where I just scrapped the masses and checked for 1:6 ratios.
But yes, it was damn long. Don't think u can solve for it on ur own without options... unless kyzoo or martoman care to find one or mao :)
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Haven't done this exam yet :S planning too soon given i've heard its hard.
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tricky i reckon.
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You could do question 4 with molar masses: find the mole of dichromate, multiply by six to find the mole of iron. Since you have the mass, you can find the molar mass of the salt.
Then, subtract the molar mass of iron to be left with the molar mass of the anion. You will find that it is equal to 2*NO3, making D the right answer.