Homework questions thread

[soccerboi]

oh i see ok thanks

[ggxoxo]

Hey guys: why do peaks [in a chromatogram] become broader with increased retention time?

AND

The organic compounds dimethyl ether (CH3OCH3) and ethanol (CH3CH2OH) have the same molecular formula, C2H6O, and the same molar mass; however they have different structural formulas.
A sample containing both ethanol and dimethyl ether is analysed by GLC using a flame ionisation detector.

If the sample has the same concentration of both chemicals, will the peaks produced have the same area? Explain your answer.

And then there's a question that says: What are the advantages of HPLC for the analysis of drugs compared to an analysis technique based on column chromatography?

But isn't HPLC a type of column chromatography? Or is column chromatography a type of chromatography by itself?

Thanks!

[dinosaur93]

An iron nail is placed in acidified potassium dischromate solutions. The chromium in the dichromate ion is reduced to Cr3+ and the ion corrodes to Fe 2+ ions. Hence what is the oxidation and reduction equation. UltimatelY the Ionic equation when the redox is combined?

[Phy124]

Firstly, as far as I know you can ignore the potassium in K₂Cr₂O₇

So we will have

Oxidation (increase in oxidation number/loss of electrons):

Fe -> Fe²⁺ + 2e^-

Reduction (decrease in oxidation number/gain of electrons):

Cr₂O₇^2- + 14 H⁺ + 6e^- -> 2 Cr³⁺ + 7 H₂O

Overall:

3 Fe + Cr₂O₇^2- + 14 H⁺ -> 3 Fe²⁺ + 2 Cr³⁺ + 7 H₂O

For states; water is liquid, Fe is solid and the rest are aqueous

Sorry if I made any mistakes did it on the iPad

Edit: I did make one, let me fix that up

Edit 2: I think it's all good now

[BambooPanda]

Hey guys, could I please get help with this question:

The molarity of an iodine solution may be determined by reacting it with a thiosulfate solution. In this procedure I2 is reduced to I(-)ions, while the S2O3(2-) ions are oxidised to S4O6(2-) ions.
An iodine solution which is known to be about 0.02 mol L1 is to be standardised. A certain mass of sodium thiosulfate is to be weighed out, and then made up to 250 mL. This solution will then be titrated against a 20 mL sample of the iodine. Theaverage titre should be approximately 20 mL for this procedure.
Calculate the mass of sodium thiosulfate required.

All I've done is derive an the ionic equation:
I2(aq)+2S203(2-)(aq) -> S4O6(2-)(aq)+2I(-)(aq)

And I'm not sure what to do after. The numbers in the brackets are charges e.g (2-), not sure how type them up properly, sorry.

Thank you!

[KOM]

Hey guys, could I please get help with this question:

The molarity of an iodine solution may be determined by reacting it with a thiosulfate solution. In this procedure I2 is reduced to I(-)ions, while the S2O3(2-) ions are oxidised to S4O6(2-) ions.
An iodine solution which is known to be about 0.02 mol L1 is to be standardised. A certain mass of sodium thiosulfate is to be weighed out, and then made up to 250 mL. This solution will then be titrated against a 20 mL sample of the iodine. Theaverage titre should be approximately 20 mL for this procedure.
Calculate the mass of sodium thiosulfate required.

All I've done is derive an the ionic equation:
I2(aq)+2S203(2-)(aq) -> S4O6(2-)(aq)+2I(-)(aq)

And I'm not sure what to do after. The numbers in the brackets are charges e.g (2-), not sure how type them up properly, sorry.

Thank you!

Okay, so you have an iodine solution which is 0.02M (mol L^-1)
And the average titre should be about 20mL
So n(I₂) = 0.02 x 0.02 = 0.0004mol

so now you know you need 0.0008 mol of sodium thiosulfate
Take note that it is the mass of Na₂S₂O₃ that you're working out, which I'm assuming from the wording of the question.

[soccerboi]

Hi im stuck on this one:
A solution of hydrogen chloride in water contains 20.15% HCl by weight and has density of 1.015g cm^-3. It is boiled at 760mmHg pressure, and has a constant boiling teperature of 110 degrees celcius. What is the molarity of this constant boiling temperature acid?
Ans is 5.610 M.

Thanks in advance

[BambooPanda]

Thanks KOM, I just realised I was finding the mass of thiosulfate only and not sodium thiosulfate!

[blank]

umm i feel really dumb but im a bit confused about oxidation numbers. what are the oxidation numbers of each element in Cu(NO3)2? 

[Phy124]

NO₃ has a charge of 1-

Therefore to have (NO₃)₂, The Cu must have a charge of 2+ (balancing the charges such that the entire compound has charge of 0)

Considering NO₃ has a charge of 1-:

O provides 3 x 2- = 6- (O is nearly always 2-)

So to get 1- for the entire NO₃ the N must be 5+

Cu = 2+
N = 5+
O = 2-

I think I wrote this a bit ambiguously, I'll try to rephrase...

[jenerator]

hey there, having trouble with this

a 1.60g sample of an organic compound was burnt in excess oxygen. 0.96g of water and 2.40g of carbon dioxide were produced. In a seperate experiment, a 2.58g sample of the compound was vapourised. the vapour occupied 1.15 L at 130 degree Celsius and 1.25 x 10^5 Pa.
Determine the molecular formula of the compound

[butene]

firstly you have to find the amount of moles of oxygen atoms
since you have the mass of h2o you can find the amount of moles of hydrogen atoms --> from there find the mass of H (tell you why later)
since you have the mass of co2 you can also find the amount of moles of carbon atoms --> then find the mass of C
the q tells you that the organic compound was burnt in EXCESS oxygen so you can't find the amount of oxygen atoms using the method above
instead of that, you know that the total mass of the organic compound is 1.60g so therefore: m(O) = 1.60 - m(H) - m(C)
from there you can calculate the empirical forumla which is CH2O
next step use the ideal gas law PV=nRT and M=n/m to find the molar mass of the actual compound
work your way there, should get C2H4O2

[thushan]

firstly you have to find the amount of moles of oxygen atoms
since you have the mass of h2o you can find the amount of moles of hydrogen atoms --> from there find the mass of H (tell you why later)
since you have the mass of co2 you can also find the amount of moles of carbon atoms --> then find the mass of C
the q tells you that the organic compound was burnt in EXCESS oxygen so you can't find the amount of oxygen atoms using the method above
instead of that, you know that the total mass of the organic compound is 1.60g so therefore: m(O) = 1.60 - m(H) - m(C)
from there you can calculate the empirical forumla which is CH2O
next step use the ideal gas law PV=nRT and M=n/m to find the molar mass of the actual compound
work your way there, should get C2H4O2

Yes! My student is awesome at chem

[mihir94]

Hi guys, i'm having trouble with his question:

Phosphorus is an essential ingredient in plant fertiliser. The phosphorus content in fertiliser can be determined as percentage by mass of P2O5.

A 3.256g sample of fertiliser is mixed with 40.0mL of deionised water and the insoluble residue os then removed using vaccum filtration. 45.0mL of 10% MgSO4.7H20 solution is added to the filtrate followed by 2M NH3 solution. A white precipitate forms. The precipitate is filtered and washed with three 5mL portions of deionised water. The final mass of the precipitate, once thoroughly dried, was 4.141g. The formula of the precpitate is known to be MgNH4PO4.6H20. Assume that the experiment was conducted at 25 degrees C and the all the phosphorus had been precipitated as MgNH4PO4.6H20.

Calculate the percentage by mass of P2O5 in the fertiliser.
Molar mass of MgNH4PO4.6H20 = 245.3g/mol and Molar mass of P2O5= 142.0 g/mol

Thanks in advance

[Phy124]

Ah, I wrote this up yesterday but wasn't going to post it because butene answered it whilst I was typing it up, but on second thought it might benefit some people so...

hey there, having trouble with this

a 1.60g sample of an organic compound was burnt in excess oxygen. 0.96g of water and 2.40g of carbon dioxide were produced. In a seperate experiment, a 2.58g sample of the compound was vapourised. the vapour occupied 1.15 L at 130 degree Celsius and 1.25 x 10^5 Pa.
Determine the molecular formula of the compound

C_xH_yO_z + O₂ -> H₂O + CO₂

n(H₂O) = m/M = 0.96/18 = 0.053 mol

n(H) = 2 n(H₂O) =  2 x 0.053 = 0.106 mol

m(H) = n x M = 0.106 x 1 = 0.106 g

n(CO₂) = m/M = 2.40/44 = 0.054 mol

n(C) = n(CO₂) = 0.054 mol

m(C) = n x M = 0.054 x 12 = 0.648 g

m(O) = m(C_xH_yO_z) – m(H) – m(C) = 1.60 – 0.106 – 0.648 = 0.846 g

You should have already established the ratio by now, from the moles above, but for the purposes of different questions when you have the masses:

Carbon      :      Hydrogen     :   Oxygen
0.648        :      0.106         :      0.846
  12                       1                      16                  <- divide by molar masses to get moles
0.054        :      0.106         :      0.053
0.053               0.053                0.053                 <- divide by lowest moles to get ratios
  1               :         2               :      1

Therefore, empirical formula is CH₂O

M(CH₂O) = 30 g/mol

Second part:

PV = nRT

n = (PV)/(RT) = (125 x 1.15)/(8.31 x 403) = 143.75/3348.93 = 0.0429 mol

M = m/n = 2.58/0.0429 = 60.106

The Molar mass is two times that of the empirical formula therefore the molecular formula is:

C₂H₄O₂