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### Topic: A few chem problems.  (Read 5207 times)

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#### usagi

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##### A few chem problems.
« on: January 20, 2009, 04:23:11 AM »
just a bit unclear on Q 33 part b). Why for Propane the number of peaks H(*) splits into is 7, but for propan-1-ol, the answers in my book say its 1? Shouldn't it be 6? since there are 5 H's as its neighbours and there are no 'shields' which block the adjacent atoms which have H's on them.

#### usagi

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##### Re: A few chem problems.
« Reply #1 on: January 20, 2009, 04:24:15 AM »
and also...

Atomic absorption spectroscopy and UV-visible spectroscopy both involve absorption of light. Both can be used to determine the amount of copper in a solution.
a) What species absorbs the light when copper nitrate is analysed by i) UV-visible spectroscopy ii) AAS?
b) Which technique would be simpliest for analysis of approximately 0.5M of copper nitrate solution? Explain.
c) How is the light of the required wavelength selected in i) UV-visible spectroscopy ii) ASS.

thanks

#### usagi

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##### Re: A few chem problems.
« Reply #2 on: January 20, 2009, 04:24:32 AM »
also Q 5... not quite sure about the table. If anyone could lend a hand that would really be helpful XD

thanks!

#### usagi

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##### Re: A few chem problems.
« Reply #3 on: January 20, 2009, 04:24:51 AM »
A sample of the energy drink High Caff was analysed for the concentration of caffeine it contained. Details on the pack indicated that it should contain 12 mg caffeine per 100 mL. A chemist decided to analyse a sample of the drink, without dilution, by HPLC. 100 mL of 10.0 $$mg mL^{-1}$$ stock standard solution of caffeine was prepared from pure caffeine tablets. The chemist decided to prepare standards of the following concentration: 5, 10, and 20 mg/100 mL caffeine. Available were 10 mL and 20 mL pipettes and 100 mL and 200 mL volumetric flasks.

a) Describe the dilutions the chemist would need to carry out in order to prepare the standard solutions using the equipments available.
b) The chemist could have used a 1 mL pipette and 100 mL volumetric flask to obtain a 10mg/100mL standard from the stock solution with only one dilution. Why is this a less accurate method?

#### usagi

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##### Re: A few chem problems.
« Reply #4 on: January 20, 2009, 04:32:58 AM »
Also

Write the ionic equation to

CH_3COOH (aq) + KOH (aq) --> CH_3OOK (aq) + H_2O (l)

How i work out ionic equation is break everything up into their ions and 'cancel' out the identical ions on the LHS and RHS of the equation.

so the equation becomes:
CH_3COO^- + H^+ + K^+ + OH^- --> CH_3COO^- + K^+ + H_2O

the CH_3COO^- and K^+ 'cancels' and we are left with H^+ (aq) + OH^- (aq) --> H_2O (l)

However my book's answers say its: CH_3COOH (aq) + OH^- (aq) --> CH_3COO^- (aq) + H_2O (l)
How do you get that? And why didn't the CH_3COO^- 'cancel' ?

Many thanks !

#### Borek

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##### Re: A few chem problems.
« Reply #5 on: January 20, 2009, 07:00:34 AM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### usagi

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##### Re: A few chem problems.
« Reply #6 on: January 20, 2009, 07:05:51 AM »
Ah thanks Borek XD

#### sjb

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##### Re: A few chem problems.
« Reply #7 on: January 20, 2009, 01:04:39 PM »
just a bit unclear on Q 33 part b). Why for Propane the number of peaks H(*) splits into is 7, but for propan-1-ol, the answers in my book say its 1? Shouldn't it be 6? since there are 5 H's as its neighbours and there are no 'shields' which block the adjacent atoms which have H's on them.

I'm not sure why it says 1, to my mind it theoretically could be 12 (dependent on resolution), do you understand why the propane proton on C2 splits into 7?

S

#### usagi

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##### Re: A few chem problems.
« Reply #8 on: January 20, 2009, 10:24:29 PM »
yeah it splits into 7 because it has 6 neighboring Hydrogens, and using the n+1 rule, 6+1 = 7. Since the hydrogens aren't being shielded off by any oxygens or whatever.

#### sjb

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##### Re: A few chem problems.
« Reply #9 on: January 21, 2009, 04:52:10 AM »
yeah it splits into 7 because it has 6 neighboring Hydrogens, and using the n+1 rule, 6+1 = 7. Since the hydrogens aren't being shielded off by any oxygens or whatever.

6 neighbouring hydrogens in identical environments, and so it splits into 7, correct. Don't discount the possibility that the answer in the back of the book is wrong