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Author Topic: VCE Chemistry Question Thread  (Read 2313544 times)  Share 

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survivor

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Re: VCE Chemistry Question Thread
« Reply #150 on: January 19, 2014, 08:11:02 pm »
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A sample of Na2CO3 contains 1.50 mole of sodium ions. Calculate the total mass of Na2CO3 in the sample.

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LOLs99

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Re: VCE Chemistry Question Thread
« Reply #151 on: January 19, 2014, 08:19:22 pm »
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A sample of Na2CO3 contains 1.50 mole of sodium ions. Calculate the total mass of Na2CO3 in the sample.

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n (Na2CO3)= 1/2 of n(Na+)
So n(Na2CO3)= 1.50mol x 0.5=0.75mol
m(Na2CO3)= 0.75mol x (23+23+12+18+18+18)= 84g

Hopefully my molar mass is correct.
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Re: VCE Chemistry Question Thread
« Reply #152 on: January 19, 2014, 08:23:12 pm »
+1
n(Na+) = m/M = 1.50
n(Na2CO3) = 1/2*n(Na+) = 0.750 mol
m(Na2CO3) = nM = 0.75*(23.0*2+12+16*3) = 79.5g to 3sigfigs

LOLs99

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Re: VCE Chemistry Question Thread
« Reply #153 on: January 19, 2014, 08:35:18 pm »
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n(Na+) = m/M = 1.50
n(Na2CO3) = 1/2*n(Na+) = 0.750 mol
m(Na2CO3) = nM = 0.75*(23.0*2+12+16*3) = 79.5g to 3sigfigs

Opps Haha thanks for reminding me my molar mass :)  oxygen is 16.
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Re: VCE Chemistry Question Thread
« Reply #154 on: January 19, 2014, 08:37:13 pm »
+1
hahah water is 18 :P

Edward Elric

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Re: VCE Chemistry Question Thread
« Reply #155 on: January 21, 2014, 02:20:45 am »
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Oxalic acid, C2H2O4 may be reacted according to the following equation.
Cr2o7^2-(aq) +8H^+(aq) + 3C2H2O4(aq) -------> 2Cr^3+(aq)+7H2O(l) + 6CO2(g)
a) write balanced half equations for the reactions occurring,
b) 20.0g of an impure sample of oxalic acid was dissolved in water to make 500.0 mL of solution. 23.0 mL of this solution was acidified with sulfuric acid, and 28.5 mL of 0.0880 M standard potassium dichromate solution was required to reach the endpoint.
i) Determine the amount (in mol) of oxalic acid in the 23.0 mL sample.
ii) Determine the percentage by mass of oxalic acid in the impure sample.

can someone do these Q and see whether i got it right as i don't have the solutions (please show all working), my answers are as follow:

a) Cr2O7^2-(aq) + 14H^+(aq) + 6e^-  ---------> 2Cr^3+(aq) + 7H2O(l)  (1)
    3C2H2O4(aq) -------> 6CO2(g) + 6H^+ + 6e^-                                     (2)
bi) 0.164 mol
ii) 73.6 %
Thanks in Advance :)

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Re: VCE Chemistry Question Thread
« Reply #156 on: January 21, 2014, 03:38:53 am »
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a) is fine
b)i) this part asks you for the amount in mol in the 23.0 mL sample, not in the entire sample, so all you need to do is this:
n(Cr2O7^2-) = cV = 0.0880*28.5*10^-3=0.002508mol
n(C2H2O4)=3*n(Cr2O7^2-) = 3*0.002508 = 0.00752 mol (correct to 3sigfigs)

ii) n(C2H2O4)total= 0.007524 * 100/23 = 0.1636 mol
m(C2H2O4)=nM=0.1636*(12*2+1*2+16*4)=14.7g
% mass = 14.7/20.0 * 100 = 73.6% (correct to 3 sigfigs)

BLACKCATT

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Re: VCE Chemistry Question Thread
« Reply #157 on: January 21, 2014, 09:42:00 am »
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ii) n(C2H2O4)total= 0.007524 * 500/23 = 0.1636 mol

just a correction, sorry

Edward Elric

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Re: VCE Chemistry Question Thread
« Reply #158 on: January 21, 2014, 02:08:06 pm »
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a) is fine
b)i) this part asks you for the amount in mol in the 23.0 mL sample, not in the entire sample, so all you need to do is this:
n(Cr2O7^2-) = cV = 0.0880*28.5*10^-3=0.002508mol
n(C2H2O4)=3*n(Cr2O7^2-) = 3*0.002508 = 0.00752 mol (correct to 3sigfigs)

ii) n(C2H2O4)total= 0.007524 * 100/23 = 0.1636 mol
m(C2H2O4)=nM=0.1636*(12*2+1*2+16*4)=14.7g
% mass = 14.7/20.0 * 100 = 73.6% (correct to 3 sigfigs)

Thank you very much Scribble :D

ETTH96

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Re: VCE Chemistry Question Thread
« Reply #159 on: January 21, 2014, 02:16:23 pm »
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How do I find the new pH?

0.100 M NaOH solution is added to a 10.0 mL aliquot of 0.100 M HCl solution

Thanks

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Re: VCE Chemistry Question Thread
« Reply #160 on: January 21, 2014, 02:18:59 pm »
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does it say how much NaOH solution has been added?
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ETTH96

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Re: VCE Chemistry Question Thread
« Reply #161 on: January 21, 2014, 02:22:36 pm »
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Nope, I've been given 4 different graphs and it asks which best represents the change in pH... How do I know which one it is?

Ive summed it down to two graphs, since its a base added to a acid the graph should go upwards.. right?

Graph 1:
EP is at 5
Highest pH is 9
Lowest pH is 1

Graph 2:
EP is at 7
Highest pH is 13
Lowest pH is 1

Sorry, if this is a bit hard to imagine..

ETTH96

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Re: VCE Chemistry Question Thread
« Reply #162 on: January 21, 2014, 02:24:51 pm »
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Oh wait, sorry it does say on the graph: 0.010 L NaOH added. SORRY
« Last Edit: January 21, 2014, 02:27:23 pm by ETTH96 »

emilyhobbes

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Re: VCE Chemistry Question Thread
« Reply #163 on: January 21, 2014, 02:43:23 pm »
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hahah, no worries!
Okay, so HCl is a strong acid and NaOH is a strong base, so this means that the end point is going to happen around ph7 and it'll be sharp

and yes, the graph should go upwards, so it sounds like graph 2 to me :)
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ETTH96

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Re: VCE Chemistry Question Thread
« Reply #164 on: January 21, 2014, 05:01:59 pm »
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Thanks emilyhobbes, really helpful! Can anyone help me with this question? Its pretty long :(

Hydrogen peroxide added with potassium permanganate...
Equation:    2MnO4(-) +5H2O2 +6H+ ==> 2Mn(2+) +8H2O +5O2


a)  The label on the bottle of the solution of hydrogen peroxide says it is 10.0%, by mass, hydrogen peroxide, and has a density of 1.19 gmL-1. Show that this data is consistent with a concentration of 3.50 M

^I've just completed this question, I did this:

1000 x 1.19 = 1190g
therefore 10% of 1190 is 119g

n(H2O2)= 119/34 = 3.50 mol
c(H2O2)=3.50/1 = 3.50 M


I STILL NEED HELP WITH THIS, PLEASE:
b)  20.00 ml of this solution is pipetted into a 250ml volumetric flask, and the volume is made up to the mark with distilled water. 25.00 ml of this solution is titrated, in the presence of dilute sulphuric acid, with a 0.129 M solution of potassium permanganate, KMnO4.
What volume of the potassium permanganate solution is required for complete reaction of 25.00 ml of the diluted sample?
« Last Edit: January 21, 2014, 06:37:03 pm by ETTH96 »