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Jason12

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Re: VCE Chemistry Question Thread
« Reply #225 on: January 25, 2014, 11:56:36 pm »
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A haemoglobin molecule contains four iron atoms and when analysed is found to contain 0.33% iron by mass. What is the molecular mass, Mr of haemoglobin?

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Re: VCE Chemistry Question Thread
« Reply #226 on: January 26, 2014, 12:13:26 am »
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iron has an atomic mass of around 55.8. so four iron atoms would have combined mass of 223.2. if we denote the molecular mass of haemoglobin as x, then 0.33/100 * x = 223.2. so x = 67636.4 g/mol. pretty big molecule.
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #227 on: January 26, 2014, 01:02:45 am »
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iron has an atomic mass of around 55.8. so four iron atoms would have combined mass of 223.2. if we denote the molecular mass of haemoglobin as x, then 0.33/100 * x = 223.2. so x = 67636.4 g/mol. pretty big molecule.

thanks but i'm having trouble trying to figure out the steps you took. I realise that the mass is 223.2g and the mol is 0.33/100? so would you use Mr=m/n?
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #228 on: January 26, 2014, 01:20:12 am »
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A solution was prepared by dissolving 4.680g of sodium chloride and 1.705g of sodium sulfate in water, so the total volume of the solution was 500ml. what is the sodium ion concentration.

so far I have done and

so what do I do from here? Am I able to add the 2 mols together then divide by 0.5L even though the substances were different?
« Last Edit: January 26, 2014, 01:59:43 am by Jason12 »
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Re: VCE Chemistry Question Thread
« Reply #229 on: January 26, 2014, 05:25:23 am »
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A solution was prepared by dissolving 4.680g of sodium chloride and 1.705g of sodium sulfate in water, so the total volume of the solution was 500ml. what is the sodium ion concentration.

so far I have done and

so what do I do from here? Am I able to add the 2 mols together then divide by 0.5L even though the substances were different?
I will continue from your step since it looks good. Just make sure u use the VCAA data booklet (download from vcaa website or ask your teacher for a nice copy).

From Na2SO4: n(Na+) = n(Na2SO4) x 2 (as there a 2 sodium atoms present in each molecule)
             = 0.0204 mol
From NaCl: n(Na+)= n(NaCl)= 0.0801 mol

So now we have all the number of moles of sodium ion from 2 solution.
Add them up. n(Na+)total= 0.0801mol+0.0204mol= 0.1005mol
Finally divide by the total volume. c(Na+)= 0.1005mol/0.500L = 0.201 M

Hope it is right. ** Yawn **
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Jason12

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Re: VCE Chemistry Question Thread
« Reply #230 on: January 26, 2014, 02:29:13 pm »
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Write balanced equations for each of the following:

Magnesium ribbon is burnt in air
Potassium metal dissolves violently in water
An aqueous sodium hydroxide solution is neutralised with  dilute aqueous sulfuric acid

Some explanation would be helpful as well for example would burnt in air mean I should add O2 to the reactants?
« Last Edit: January 26, 2014, 02:30:55 pm by Jason12 »
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Re: VCE Chemistry Question Thread
« Reply #231 on: January 26, 2014, 02:31:39 pm »
+1
Write balanced equations for each of the following:

Magnesium ribbon is burnt in air
Potassium metal dissolves violently in water
An aqueous sodium hydroxide solution is neutralised with  dilute aqueous sulfuric acid

(a) 2Mg(s) + O2(g) ---> 2MgO(s)

(b) K(s) + H2O(l) ---> KO(aq) + H2(g)

(c) 2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l)

I could be wrong, but that's my try ^^. Clarification would be great :)

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Re: VCE Chemistry Question Thread
« Reply #232 on: January 26, 2014, 02:48:58 pm »
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And also an aqueous dilute nitric acid solution reacts with solid zinc carbonate
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Re: VCE Chemistry Question Thread
« Reply #233 on: January 26, 2014, 03:08:47 pm »
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2HNO3(aq) + ZnCO3(s) --->  Zn(NO3)2 + H2O(l) + CO2(g)
 
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Re: VCE Chemistry Question Thread
« Reply #234 on: January 26, 2014, 04:28:36 pm »
+1
(a) 2Mg(s) + O2(g) ---> 2MgO(s)

(b) K(s) + H2O(l) ---> KO(aq) + H2(g)

(c) 2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l)

I could be wrong, but that's my try ^^. Clarification would be great :)


2 K(s) + 2 H2O(l) ---> 2 KOH (aq) + H2 (g)

is what you mean.
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Re: VCE Chemistry Question Thread
« Reply #235 on: January 26, 2014, 05:19:31 pm »
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2 K(s) + 2 H2O(l) ---> 2 KOH (aq) + H2 (g)

is what you mean.

I see :) Thanks thushan.

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Re: VCE Chemistry Question Thread
« Reply #236 on: January 27, 2014, 12:33:39 am »
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Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..
 
Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)


So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
      S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)


I may have done something wrong, who knows.
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #237 on: January 27, 2014, 12:50:25 am »
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Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..
 
Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)


So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
      S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)


I may have done something wrong, who knows.

(ii) should have a coefficient of 1 for manganese dioxide
(iii) should therefore be slightly different
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Re: VCE Chemistry Question Thread
« Reply #238 on: January 27, 2014, 10:08:57 am »
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Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..
 
Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)


So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
      S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)


I may have done something wrong, who knows.


8H+(aq) + 2MnO2(s) +4e  ---> 2Mn2+(aq) + 4H2O(l)

S(s) +2H2O(l) ---> SO2(g) + 4H+(aq) + 4e

Balanced equation:
4H+(aq) + 2MnO2(s)  ---> 2Mn2+(aq) + 2H2O(l)

S(s)  ---> SO2(g)

Balanced equation:
S(s) + 2MnO2(s) + 4H+(aq) ---> 2Mn2+(aq) + 2H2O(l) + SO2(g)


lznxl- why is the coefficient of manganese dioxide 1? :/

T-Infinite

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Re: VCE Chemistry Question Thread
« Reply #239 on: January 27, 2014, 10:17:01 am »
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8H+(aq) + 2MnO2(s) +4e  ---> 2Mn2+(aq) + 4H2O(l)

S(s) +2H2O(l) ---> SO2(g) + 4H+(aq) + 4e

Balanced equation:
4H+(aq) + 2MnO2(s)  ---> 2Mn2+(aq) + 2H2O(l)

S(s)  ---> SO2(g)

Balanced equation:
S(s) + 2MnO2(s) + 4H+(aq) ---> 2Mn2+(aq) + 2H2O(l) + SO2(g)


lznxl- why is the coefficient of manganese dioxide 1? :/
Just wondering, what happened to the 8H+ on the left and the 4H+ on the right? Did they cancel out leaving only 4H+ on the left for the final balanced equation? o-o
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