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October 31, 2025, 05:26:20 am

Author Topic: Stankovic123's chem q's  (Read 72532 times)  Share 

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zvezda

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Re: Stankovic123's chem q's
« Reply #105 on: April 28, 2013, 07:00:32 pm »
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You're right that is complicated.

With redox titrations, is there a reason why an extra titration is carried out on iodine with sodium thiosulfphate after an initial titration that involved iodine ions?
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Limista

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Re: Stankovic123's chem q's
« Reply #106 on: April 28, 2013, 07:45:45 pm »
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You're right that is complicated.

With redox titrations, is there a reason why an extra titration is carried out on iodine with sodium thiosulfphate after an initial titration that involved iodine ions?

I think you are referring to a back titration here. Just because you carry out a redox titration, it does not mean it has to be a back titration. We need to use a back titration for other reasons, such as when the reactants are volatile or if the strength of the acids/bases are weak.
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jgoudie

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Re: Stankovic123's chem q's
« Reply #107 on: April 28, 2013, 08:49:52 pm »
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The reason for the iodine titration is due to sodium thiosulphate not having a colour change, or at least one that is quick.  Iodine and starch give you a sharp end point, thus you can easily calculate the amount of iodine, then, though stoic find thiosulphate and everything else you might need to.  As stated, back titration can be used for a number of reasons, most notably when you do not have an easy to see end point between your analyte.


You're right that is complicated.

With redox titrations, is there a reason why an extra titration is carried out on iodine with sodium thiosulfphate after an initial titration that involved iodine ions?
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zvezda

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Re: Stankovic123's chem q's
« Reply #108 on: May 28, 2013, 09:30:00 pm »
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Hey,
Is there a reason why an exact/certain level of energy is required to promote and electron to a higher energy level?
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Aurelian

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Re: Stankovic123's chem q's
« Reply #109 on: May 28, 2013, 09:36:25 pm »
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Hey,
Is there a reason why an exact/certain level of energy is required to promote and electron to a higher energy level?

Yes, but you're unlikely to find a satisfying explanation at year 12 level. Do you do physics? If so, some preliminary light will be shed when you do the "light and matter" area of study =)
« Last Edit: May 28, 2013, 09:38:31 pm by Aurelian »
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lzxnl

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Re: Stankovic123's chem q's
« Reply #110 on: May 28, 2013, 09:39:22 pm »
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I'll give you an elementary explanation. It was soon discovered that all matter has wave-like properties as well. Imagine a standing wave set up in a pipe. Only certain wavelengths can fit into the pipe; these are the harmonics that we hear in brass and woodwind instruments. As a wave's energy is related to the wavelength and frequency, if an electron is a wave and can only have specific wavelengths, its energy can also only take specific values. Therefore, energy promotions are quite specific.
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zvezda

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Re: Stankovic123's chem q's
« Reply #111 on: May 28, 2013, 10:02:26 pm »
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Yes, but you're unlikely to find a satisfying explanation at year 12 level. Do you do physics? If so, some preliminary light will be shed when you do the "light and matter" area of study =)

Nah I dont do physics, but thanks anyway.

I'll give you an elementary explanation. It was soon discovered that all matter has wave-like properties as well. Imagine a standing wave set up in a pipe. Only certain wavelengths can fit into the pipe; these are the harmonics that we hear in brass and woodwind instruments. As a wave's energy is related to the wavelength and frequency, if an electron is a wave and can only have specific wavelengths, its energy can also only take specific values. Therefore, energy promotions are quite specific.

Ah yeah. Pretty elementary but I suppose it's enough for now. Cheers
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zvezda

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Re: Stankovic123's chem q's
« Reply #112 on: June 01, 2013, 09:05:01 pm »
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Hey,
With uv spectroscopy, what is the light that leaves the sample solution and enters the light detector?
Is it the energy that has been emitted after having been absorbed by the sample?
Thanks in advance
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Mao

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Re: Stankovic123's chem q's
« Reply #113 on: June 02, 2013, 06:50:27 pm »
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Hey,
With uv spectroscopy, what is the light that leaves the sample solution and enters the light detector?
Is it the energy that has been emitted after having been absorbed by the sample?
Thanks in advance

To a very small extent. It mostly the original beam which did not get absorbed.

E.g. light at a illuminance is shown at the cell, 30% gets absorbed, the light that enters the detector would then have an illuminance of . The figure of 70% is the 'transmittance'. This makes up most of the light entering the detector.

Note that light is not necessarily 'emitted' by the molecules after absorption. The absorbed light can be dissipated via other pathways that do not emit light (e.g. via bending/vibrations of the molecules), or emit light at a different frequency (e.g. fluorescence/phosphorescence).

Sometimes, light is emitted by the excited molecule at the same frequency. However, because light is emitted in all directions, it typically contributes very little to the total transmittance. Furthermore, even if the emitted light is significant, the linearity of the absorbance-vs-concentration relationship would be preserved, so our usual calibration curve method still works. Note that the measured transmittance includes the effect of light emitted at the same frequency.
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zvezda

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Re: Stankovic123's chem q's
« Reply #114 on: June 03, 2013, 09:28:37 pm »
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To a very small extent. It mostly the original beam which did not get absorbed.

E.g. light at a illuminance is shown at the cell, 30% gets absorbed, the light that enters the detector would then have an illuminance of . The figure of 70% is the 'transmittance'. This makes up most of the light entering the detector.

Note that light is not necessarily 'emitted' by the molecules after absorption. The absorbed light can be dissipated via other pathways that do not emit light (e.g. via bending/vibrations of the molecules), or emit light at a different frequency (e.g. fluorescence/phosphorescence).

Sometimes, light is emitted by the excited molecule at the same frequency. However, because light is emitted in all directions, it typically contributes very little to the total transmittance. Furthermore, even if the emitted light is significant, the linearity of the absorbance-vs-concentration relationship would be preserved, so our usual calibration curve method still works. Note that the measured transmittance includes the effect of light emitted at the same frequency.

So how would this relationship be preserved?
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zvezda

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Re: Stankovic123's chem q's
« Reply #115 on: June 03, 2013, 09:28:54 pm »
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Sulfur dioxide, SO2, is a serious air pollutant, and its level in the atmosphere therefore requires close monitoring.
In one particular analysis, 10 000 L of air (at 20oC and 1 atm pressure) was bubbled through 100 mL of 0.037 34 mol L 1 KMnO4 solution. The SO2 in the air reacted with the MnO4 ions present according to the following equation:
5SO2(g) 2MnO4 (aq) 2H2O(l)
5SO42 (aq) 2Mn2 (aq) 4H (aq)

After the completion of this step, the remaining MnO4 was titrated with Fe2 solution, according to the following equation:
5Fe2(aq)  MnO4(aq)  8H(aq)
5Fe3(aq)  Mn2(aq)  4H2O(l)
Assume 31.62 mL of 0.1843 mol L1 of Fe2 solu- tion was required to react with all of the remaining KMnO4.
Use these data to calculate:

(h) the level, in ppm, of SO2 in the air sample.

Hey, I'm having a small issue with part h,
I've calculated the concentration of so2 in ppm in two ways: mg/L and L/L x 10^6.
The issue is, the 2 final values are different. Is this supposed to be happening?
Thanks in advance
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zvezda

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Re: Stankovic123's chem q's
« Reply #116 on: June 15, 2013, 01:46:55 pm »
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This is a question from the 2009 VCAA unit 3 exam. It asked to explain why the sodium salt of aspirin is more water-soluble than aspirin.
I had a look at the answer, and to me it doesnt seem like its enough. I mean, aspirin can form hydrogen bonds with water cant it? Saying that the ion-dipole bonds would be stronger than the hydrogen bonds wouldnt really justify why the salt is soluble whereas the aspirin is not?
Thoughts?
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psyxwar

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Re: Stankovic123's chem q's
« Reply #117 on: June 15, 2013, 02:01:57 pm »
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This is a question from the 2009 VCAA unit 3 exam. It asked to explain why the sodium salt of aspirin is more water-soluble than aspirin.
I had a look at the answer, and to me it doesnt seem like its enough. I mean, aspirin can form hydrogen bonds with water cant it? Saying that the ion-dipole bonds would be stronger than the hydrogen bonds wouldnt really justify why the salt is soluble whereas the aspirin is not?
Thoughts?
Well, sure it might be able to hydrogen bond, but if the salt is dissociating into ions with formal charge then obviously the forces with water (ion-dipole) would be stronger than that from hydrogen bonding. The question isn't asking about why the salt is soluble; it's asking why its more soluble. Hence, the stronger forces of attraction with water would definitely make it more soluble relative to aspirin, which is what the question is asking for.
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zvezda

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Re: Stankovic123's chem q's
« Reply #118 on: June 15, 2013, 02:25:57 pm »
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Well, sure it might be able to hydrogen bond, but if the salt is dissociating into ions with formal charge then obviously the forces with water (ion-dipole) would be stronger than that from hydrogen bonding. The question isn't asking about why the salt is soluble; it's asking why its more soluble. Hence, the stronger forces of attraction with water would definitely make it more soluble relative to aspirin, which is what the question is asking for.

That's a good point. I guess that makes more sense then. Somehow ignored the more part of the question. Thanks for the help
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zvezda

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Re: Stankovic123's chem q's
« Reply #119 on: June 15, 2013, 02:30:14 pm »
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Another question,
Why do we have both UV/vis and AAS. Aren't they effectively serving the same purpose in the same way? They both use energy from basically the same area of the electromagnetic spectrum, just that UV uses UV waves as well. So I guess, aside from the substances that don't absorb visible light waves (thus calling for the use of UV instead of AAS), why do we have these 2?
I know that AAS can analyze very small amounts of substance, if that contributes at all to this.
Help is much appreciated
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