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November 01, 2025, 11:36:58 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448755 times)  Share 

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thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #735 on: April 19, 2013, 09:41:15 am »
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Should we say 2,2,2-trichloroethan-1,1-diol or 2,2,2-trichloroethane-1,1-diol? We normally have ethanol...without the extra e.

Keep the 'e' - its basically because it rolls better off the tongue.
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Re: Chemistry 3/4 2013 Thread
« Reply #736 on: April 19, 2013, 10:31:26 pm »
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Can some one please help me with these spectroscopy and chromatography questions.

If an individual suspected of taking the drug heroin and  overdosed on it has his blood sample run through the HPLC resulting in over 100  peaks present in the chromatgram. Why are there so many peaks? 

   
What wavelength should be selected to generate a calibiration curve of the relative absorbances of a set standard of solutions of known fluroscien solution using UV VIS.  (fluroescin-glows green in very low concentrations). 


If you have a ppm of 0.60( you found out) in which a 0.520 g sample of a substanece was dissolved in 100 ml of water what is the percentage of the sample in the the 0.520 g. How would you work this out? 

Thank you in advance
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #737 on: April 20, 2013, 11:00:59 pm »
+1
Lithium metal is the strongest reductant there is, so if connected in a redox cell with something it can reduce, it will produce a larger voltage than any other reductant. It also has a small molar mass, allowing for more lithium atoms per unit mass. A battery is a galvanic cell, except contained in the one object. Rechargeable batteries utilize electrolysis to reverse the redox reactions, allowing for reuse.
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Re: Chemistry 3/4 2013 Thread
« Reply #738 on: April 21, 2013, 11:39:02 am »
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Actually...I was talking about a regular lithium battery, not a lithium ion battery. In a regular lithium battery, the aluminium or copper electrodes act as oxidants to oxidise the lithium metal, thus completing the redox cell.
An actual lithium ion battery is slightly different. It has lithium compounds at both the anode and cathode. Read http://en.wikipedia.org/wiki/Lithium-ion_battery for more info.
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Re: Chemistry 3/4 2013 Thread
« Reply #739 on: April 21, 2013, 11:38:25 pm »
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Can some one please help me with these spectroscopy and chromatography questions.

If an individual suspected of taking the drug heroin and  overdosed on it has his blood sample run through the HPLC resulting in over 100  peaks present in the chromatgram. Why are there so many peaks? 

When does a peak occur? When components in the sample pass through the detector, which is at the end of the column. The individual overdosed on heroin, meaning that he took more than he should have, which means he probably had exposure to more than the recommended number of components - he had exposure to at least 100 components, which represents the 100+ peaks on his chromatogram.


If you have a ppm of 0.60( you found out) in which a 0.520 g sample of a substanece was dissolved in 100 ml of water what is the percentage of the sample in the the 0.520 g. How would you work this out? 

C(substance) = 0.60 ppm

10^6g = 0.60g
0.520g = 3.1*10^(-7)g
% = 6.0*10^(-5)% (w/w)

Thank you in advance
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Patches

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Re: Chemistry 3/4 2013 Thread
« Reply #740 on: April 22, 2013, 11:46:13 am »
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Quote
If an individual suspected of taking the drug heroin and  overdosed on it has his blood sample run through the HPLC resulting in over 100  peaks present in the chromatgram. Why are there so many peaks? 
When does a peak occur? When components in the sample pass through the detector, which is at the end of the column. The individual overdosed on heroin, meaning that he took more than he should have, which means he probably had exposure to more than the recommended number of components - he had exposure to at least 100 components, which represents the 100+ peaks on his chromatogram.

Surely blood already contains a lot of compounds even when heroin isn't present? It doesn't say whether the heroin is pure or not - surely it will be responsible for one (or maybe a few) peaks, but the majority of the 100 are going to be from things normally in the blood.

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Re: Chemistry 3/4 2013 Thread
« Reply #741 on: April 22, 2013, 06:46:42 pm »
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If an individual suspected of taking the drug heroin and  overdosed on it has his blood sample run through the HPLC resulting in over 100  peaks present in the chromatgram. Why are there so many peaks? 

I think the reason why there are so many peaks is because the heroin tablet and the blood contained over 100 different substances. I don't think the dose of heroin can change the amount of peaks, just the area under the peaks.

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Re: Chemistry 3/4 2013 Thread
« Reply #742 on: April 22, 2013, 08:31:35 pm »
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I think the reason why there are so many peaks is because the heroin tablet and the blood contained over 100 different substances. I don't think the dose of heroin can change the amount of peaks, just the area under the peaks.


Thats what i thought.

BTW does anyone one which out of these 4 is most strongly to be deflected from a mass spectroemetor

CH4+
CH2^2+
CH3+
CH4^2+

thanks in advance

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Re: Chemistry 3/4 2013 Thread
« Reply #743 on: April 22, 2013, 09:37:43 pm »
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This will make more sense if you do physics.
Basically, what we have here is a magnetic field that acts on charged molecular ions. A uniform magnetic field will cause these charges to move in a circle as the magnetic force is always perpendicular to the the motion of the ions.
From circular motion formula, force = mv^2/r
But force also equals qvB from the magnetic force equation
mv^2/r=qvB
r=mv/qB

Here, the magnetic field strength B remains constant. The v here is simply how fast the molecular ion was moving before it hit the magnetic field (forces perpendicular to direction of motion cannot affect kinetic energies). Therefore, m and q are the only variables which affect r, the circle radius that the ion travels in and hence the deflection. It is this mass to charge ratio that is measured in mass spectrometry.
Out of your four, CH4 + has the largest mass and the smallest charge. Therefore it will be deflected further.

Think of it this way if you like. Magnetic force = qvB. If you have a smaller q, the magnetic force is weaker. Therefore you get a weaker centripetal acceleration. If you decrease v^2/r while keeping v constant, r must increase.

This will make no sense if you don't do physics, so I apologise if that is the case. If you don't do physics, simply remember that the mass divided by the charge determines how much the ion is deflected. Mass divided by charge NOT the other way around :D
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Re: Chemistry 3/4 2013 Thread
« Reply #744 on: April 22, 2013, 11:54:26 pm »
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Anyone know how i should study for a NMR spectroscopy sac?

Patches

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Re: Chemistry 3/4 2013 Thread
« Reply #745 on: April 23, 2013, 09:04:27 am »
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This will make more sense if you do physics.
Basically, what we have here is a magnetic field that acts on charged molecular ions. A uniform magnetic field will cause these charges to move in a circle as the magnetic force is always perpendicular to the the motion of the ions.
From circular motion formula, force = mv^2/r
But force also equals qvB from the magnetic force equation
mv^2/r=qvB
r=mv/qB

Here, the magnetic field strength B remains constant. The v here is simply how fast the molecular ion was moving before it hit the magnetic field (forces perpendicular to direction of motion cannot affect kinetic energies). Therefore, m and q are the only variables which affect r, the circle radius that the ion travels in and hence the deflection. It is this mass to charge ratio that is measured in mass spectrometry.
Out of your four, CH4 + has the largest mass and the smallest charge. Therefore it will be deflected further.

Think of it this way if you like. Magnetic force = qvB. If you have a smaller q, the magnetic force is weaker. Therefore you get a weaker centripetal acceleration. If you decrease v^2/r while keeping v constant, r must increase.

This will make no sense if you don't do physics, so I apologise if that is the case. If you don't do physics, simply remember that the mass divided by the charge determines how much the ion is deflected. Mass divided by charge NOT the other way around :D

Our teacher told us that the heavier particle will be deflected the least because it has more momentum - is it safe to assume he's wrong?

thushan

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Re: Chemistry 3/4 2013 Thread
« Reply #746 on: April 23, 2013, 02:39:20 pm »
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Nah Patches, your teacher is right. nliu basically said the radius that a particle would travel in a circle if exposed to a magnetic field is mass/charge.

A greater radius means a lower deflection (a bigger circle means less sharp arcs) - think about it visually.

r = mv/qB

So greater m implies greater r, meaning less deflection. So you're right Patches.

Greater q implies smaller r, meaning greater deflection.

But don't worry about the r = mv/qB. Just think of it in terms of what your teacher said Patches.

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Btw the answer would be the particle with the highest charge/mass (lowest mass/charge) ratio.
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Alwin

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Re: Chemistry 3/4 2013 Thread
« Reply #747 on: April 23, 2013, 05:02:43 pm »
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Here, the magnetic field strength B remains constant. The v here is simply how fast the molecular ion was moving before it hit the magnetic field (forces perpendicular to direction of motion cannot affect kinetic energies). Therefore, m and q are the only variables which affect r, the circle radius that the ion travels in and hence the deflection. It is this mass to charge ratio that is measured in mass spectrometry.
Out of your four, CH4 + has the largest mass and the smallest charge. Therefore it will be deflected further.

Sorry, but wrong. What you sir found was the the particle with the LARGEST radius. You want the SMALLEST radius since more deflection! The correct answer is the particle with the lowest mass divided by charge.

Another way, just to visualise it is say a a canon ball and a water gun are shot in to a cross wind. Obviously the canon ball will be deflected much less than the water droplets, because it is much much heavier. Dividing by charge allows you to factor in the effects of interactions between the field and the charge.
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Re: Chemistry 3/4 2013 Thread
« Reply #748 on: April 23, 2013, 06:21:39 pm »
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In UV-visible spectroscopy, what part of the substance actually absorbs the electromagnetic radiation? My textbook implies that it is the electrons, but my teacher told us it's actually the bonds. I'm a tad confused. :S
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Re: Chemistry 3/4 2013 Thread
« Reply #749 on: April 23, 2013, 06:28:18 pm »
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In UV-visible spectroscopy, what part of the substance actually absorbs the electromagnetic radiation? My textbook implies that it is the electrons, but my teacher told us it's actually the bonds. I'm a tad confused. :S

Don't the electrons form the bonds?