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March 29, 2024, 12:22:07 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313529 times)  Share 

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nhmn0301

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Re: VCE Chemistry Question Thread
« Reply #255 on: January 31, 2014, 08:11:25 am »
+1
Can someone please clarify that i've understood this correctly:

For the equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g),
we know that n(Mg)=0.0551 and n(HCl)=0.0600.

Since the Mg is in excess, the amount reacting would follow the mole ratio. This means, 0.03mol of Mg would react with 0.0600 mol of HCl. This would produce 0.09mol of MgCl2 and H2 gas.
Looking at the mole ratio: you have 1:2:1:1
You are right at the point Mg is the excess reactant, hence the calculation will base on the mole of HCl. According to the mole ratio above, if you have 0.0600 mole of HCl, Mg will be halved that mole amount and equal 0.0300 mole. Other products also have the same mole ratio as Mg (1:1:1), therefore, MgCl2 and H2 will have the 0.0300 as well. 
Remember that determining mole is based on the mole ratio on the equation you got. I think you are confused with the Conservation of Mass theory, which means the MASS of reactants must equal the mass of products. Note that this is mole, not mass.
Hope this helps!
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psyxwar

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Re: VCE Chemistry Question Thread
« Reply #256 on: February 01, 2014, 10:11:53 am »
0
Can someone please clarify that i've understood this correctly:

For the equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g),
we know that n(Mg)=0.0551 and n(HCl)=0.0600.

Since the Mg is in excess, the amount reacting would follow the mole ratio. This means, 0.03mol of Mg would react with 0.0600 mol of HCl. This would produce 0.09mol of MgCl2 and H2 gas.
correct, except your sig figs are off.
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Sanguinne

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Re: VCE Chemistry Question Thread
« Reply #257 on: February 01, 2014, 11:57:03 am »
0
What are the states for organic compound such as esters?
Would they be aqueous or solid or gas or liquid?

For example, if i had a question which asked me to write the equation of the formation of ethyl butanoate, i get
                                   H2SO4
CH3CH2OH + CH3CH2CH2COOH ------> CH3CH2CH2COOCH2CH3 + H20(l)

what are the states for this organic compounds?
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brightsky

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Re: VCE Chemistry Question Thread
« Reply #258 on: February 01, 2014, 11:59:20 am »
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Usually liquid.
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darklight

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Re: VCE Chemistry Question Thread
« Reply #259 on: February 01, 2014, 01:01:30 pm »
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I've heard that states aren't that important in organic chem. Is this true?
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Edward21

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Re: VCE Chemistry Question Thread
« Reply #260 on: February 01, 2014, 02:07:32 pm »
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I've heard that states aren't that important in organic chem. Is this true?
IF you've got a combustion equation, you'd put what you're oxidising in the (l) state. States still need to be there though! Compare the states that you've put down, with what's in the answer. It's confusing when we ferment glucose and get CO2 and ethanol, because some reason that both the glucose and ethanol are (l), the ethanol only is (aq) or none are (l).. :\ i've seen this vary from resource to resource unfortunately..
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #261 on: February 01, 2014, 03:19:30 pm »
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Can someone explain how to:

Calculate the mass, in mg, of manganese in the steel sample

in the attached question?

This is SA Q. 2bii) of the 2008 Chemistry Exam (if the image doesn't work)
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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #262 on: February 01, 2014, 04:37:06 pm »
+1
From the graph:
[Mn] = 35 (however no units are given)

Because no units are given we cant really complete the whole question but lets do what we can.

This [Mn] being 35 is in a diluted sample.  So we need to find the concentration of Mn in the undiluted sample.

C1V1=C2V2
C1= ?
V1= 25ml
C2= 35
V2= 100ml

C1=35*100/25
    = 140
So we have a concentration of the undiluted sample in the 1L being 140.  My guess is you cut off the x-axis label with the units of concentration. We need this to complete the question.

Can someone explain how to:

Calculate the mass, in mg, of manganese in the steel sample

in the attached question?

This is SA Q. 2bii) of the 2008 Chemistry Exam (if the image doesn't work)
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #263 on: February 01, 2014, 06:05:03 pm »
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My guess is you cut off the x-axis label with the units of concentration. We need this to complete the question.
I'm so sorry!! Yes I did. The unit of measure is the concentration of MnO4- (aq) (mgL-1)
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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #264 on: February 01, 2014, 10:01:59 pm »
+1
Okay, so where were we...  (please note i have changed my tune a little from my last post, the graph tells us [MnO4-] not [Mn]

140mg/L  this is the mass of MnO4- in the undiluted solution

You are looking for only the Mn not the MnO4-.  so: %Mn in  MnO4- = 46.17%.

mass of Mn in 140mg of    MnO4-    = 140*0.4617
                     = 64.64mg
                     = 0.06464g

%Mn in sample.     = Mass of Mn/Mass of sample*100
            = 0.06464/13.936*100
            = 0.46%


I'm so sorry!! Yes I did. The unit of measure is the concentration of MnO4- (aq) (mgL-1)
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EFPBH

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Re: VCE Chemistry Question Thread
« Reply #265 on: February 02, 2014, 12:17:13 am »
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Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)

Edward21

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Re: VCE Chemistry Question Thread
« Reply #266 on: February 02, 2014, 12:36:27 am »
+2
Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)

MgCO3(s) + 2HCl(aq) --> CO2(g) + H2O(l) + MgCl2(aq)

If pV=nRT

n(CO2)=(101.3kPa x 0.714L)/(8.31 x (22+273)K)=0.0296mol
n(MgCO3)=n(CO2)=0.0295mol
m(MgCO3)=nM=0.0295mol x 84.3gmol=2.49g

%MgCO3=(2.49g)/(3.50g) x 100% = 71.1% purity  :)
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bucklr

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Re: VCE Chemistry Question Thread
« Reply #267 on: February 02, 2014, 12:42:14 am »
+2
Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)
using - MgCO3(s)+2HCL(aq) -> MgCl(aq)+H2O(l)+CO2(g)

0.714L of CO(2) produced
22.0oC -> convert to Kelvin by adding 273 -> 295oK

use gas equation to find mol of CO2

1=(0.714x101.3)/(8.31xnx295)
n=0.0295 mol of CO2

use balanced chemical equation to find mol of MgCO2

therefore... n(MgCO3)=0.0295mol
find mass of MgCO3 (using periodic table molar mass of Mg=24.3, C=12.0, O=32.0)
therefore... 0.0295x(24.3+12.0+48.0)=2.49g
find percentage by mass
100x2.49/3.50=71.1%(w/w)
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #268 on: February 02, 2014, 08:52:35 pm »
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If I have 1.305g of Na2CO3 and put this in a 250.0mL flask (and filled it up to the mark w/ deionised water) does this mean there is 1.305g of NaCO3 per 250mL?

Seems pretty straight forward but I just want to make sure :p
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jgoudie

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Re: VCE Chemistry Question Thread
« Reply #269 on: February 02, 2014, 09:00:36 pm »
+1
Yep, and you would have 5.22g per 1L,  Provided that is is pure Na2CO3


If I have 1.305g of Na2CO3 and put this in a 250.0mL flask (and filled it up to the mark w/ deionised water) does this mean there is 1.305g of NaCO3 per 250mL?

Seems pretty straight forward but I just want to make sure :p
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