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July 23, 2025, 01:10:16 am
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Author Topic: Andiio's Chem Questions!  (Read 6808 times)  Share 

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samiira

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Re: Andiio's Chem Questions!
« Reply #45 on: March 03, 2011, 10:44:27 pm »
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moles = 0.00050 g / 176 g/mol =2.8 x 10^-6

2.8 x 10^-6 x 1000 = 0.0028 mol/L of C6H8O6

If we take 50.0 mL of this solution there are :

0.050 L x 0.0028 mol/L = 0.00014 moles

Moles I2 needed = 0.00014

V of I2 = 0.00014 / 0.0100 M =0.014 L = 14 mL
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Andiio

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Re: Andiio's Chem Questions!
« Reply #46 on: March 03, 2011, 11:08:06 pm »
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Quote from: samiira on March 03, 2011, 10:44:27 pm
moles = 0.00050 g / 176 g/mol =2.8 x 10^-6

2.8 x 10^-6 x 1000 = 0.0028 mol/L of C6H8O6

If we take 50.0 mL of this solution there are :

0.050 L x 0.0028 mol/L = 0.00014 moles

Moles I2 needed = 0.00014

V of I2 = 0.00014 / 0.0100 M =0.014 L = 14 mL


Mm, but I'm just not sure why we take 50.0 mL? o_O
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Andiio

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Re: Andiio's Chem Questions!
« Reply #47 on: March 15, 2011, 08:10:04 pm »
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Could anyone please help clarify this simple Q?

Q:  A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus:
a) The pipette used to deliver the aliquot of sodium carbonate solution
b) The burette.

For part a), what I reasoned was that, the presence of water would lower the concentration of sodium carbonate solution, and as a result, the V(HCl) needed in the titration to reach the equivalence point would be lower, thus yielding a lower titre - V(HCl). And because C = n/V, since V(HCl) is lower, it would yield a higher concentration. I am not really sure if this is correct as I was thinking that maybe the student would have delivered a lower amount of HCl (and thus a lower amount of moles of HCl) in the pipette as there would be water 'taking up' a few mL's of space.

For part b), I wrote that the water would dilute the HCl (but with no effect on the amount of moles of HCl) which would then give a lower concentration of HCl.

Thanks!
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luken93

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Re: Andiio's Chem Questions!
« Reply #48 on: March 15, 2011, 08:18:35 pm »
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a) If the pipette was washed with water, then a smaller amount of Sodium Carbonate would be delivered. As a result, it would require less HCl to neutralise the NaOH. Because c = n/V, then a lower V reading will result in a higher concentration of HCl than there actually is.

b) If the burette has some water left over, then more HCl will be delivered to neutralise the NaOH. Because c = n/V, a higher V reading will result in a lower concentration than there actually is.

I haven't read your answers btw, but hopefully they match up :)

It always works better with formulas :P
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Andiio

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Re: Andiio's Chem Questions!
« Reply #49 on: April 25, 2011, 04:57:51 pm »
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Can UV-Visible spec be used to analyse metal ions at all? (Transition metals?)

How would one analyse a colourless compound using UV-Vis? (Is it merely to convert the colourless compound into a coloured one?)

What can samples be held in, apart from quartz? (UV-Vis)

Also, do we have to know about the double-beam spectrophotometer?

Thanks!
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Greatness

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Re: Andiio's Chem Questions!
« Reply #50 on: April 25, 2011, 08:22:49 pm »
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Can UV-Visible spec be used to analyse metal ions at all? (Transition metals?)
Well UV analyses molecules so i dont think you would be able to analyse metals.

How would one analyse a colourless compound using UV-Vis?
You would do it the same way, you would just have to compare the absorption spectrum with a known compound.

Also, do we have to know about the double-beam spectrophotometer?
I doubt it.
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Water

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Re: Andiio's Chem Questions!
« Reply #51 on: April 25, 2011, 08:28:53 pm »
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UV Vis can be used to analyze transition metals
« Last Edit: April 25, 2011, 08:33:09 pm by Water »
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Greatness

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Re: Andiio's Chem Questions!
« Reply #52 on: April 25, 2011, 08:41:15 pm »
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Quote from: Water on April 25, 2011, 08:28:53 pm
UV Vis can be used to analyze transition metals
Really? Link me pls :D
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Andiio

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Re: Andiio's Chem Questions!
« Reply #53 on: April 25, 2011, 08:43:16 pm »
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Quote from: swarley on April 25, 2011, 08:22:49 pm
Can UV-Visible spec be used to analyse metal ions at all? (Transition metals?)
Well UV analyses molecules so i dont think you would be able to analyse metals.

How would one analyse a colourless compound using UV-Vis?
You would do it the same way, you would just have to compare the absorption spectrum with a known compound.

Also, do we have to know about the double-beam spectrophotometer?
I doubt it.

Yep as Water said, UV-Vis spec can be used to analyse transition metals; but can't analyse metal ions? What's the diff/significance? :\

I agree with you; analysing colourless compounds quantitatively should be fine, but how about qualitative? :\ You'd have to change it to a coloured compound, wouldn't you?
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Water

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Re: Andiio's Chem Questions!
« Reply #54 on: April 25, 2011, 08:54:38 pm »
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Quote
Yep as Water said, UV-Vis spec can be used to analyse transition metals; but can't analyse metal ions? What's the diff/significance? :\


No no. Sorry, you can find out metal ions, through UV VIS.


Sorry Andio..when transition metals go in solution, they automatically become ionized. I was working on that basis.


@swarley you react the metal ions with another component, and you work from there :)
« Last Edit: April 25, 2011, 08:57:38 pm by Water »
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thushan

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Re: Andiio's Chem Questions!
« Reply #55 on: April 26, 2011, 03:53:20 pm »
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I'm not entirely sure here, but I think it is possible to use UV-vis with metal ions, but we would use AAS, because it is much more accurate and sensitive.
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luken93

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Re: Andiio's Chem Questions!
« Reply #56 on: April 26, 2011, 04:48:03 pm »
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Mao answered this a while ago, I'm trying to find it now..
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luken93

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Re: Andiio's Chem Questions!
« Reply #57 on: April 26, 2011, 04:49:33 pm »
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Found it, have a look at this thread and see if it makes anymore sense:

http://vce.atarnotes.com/forum/index.php/topic,39634.msg415664.html#msg415664
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