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November 01, 2025, 11:46:09 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448756 times)  Share 

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Mr Keshy

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Re: Chemistry 3/4 2013 Thread
« Reply #255 on: January 25, 2013, 12:06:02 am »
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I said this ages ago!!! Thush is stealing my thunder :P

No one has to be in charge as such :) We can all just help out when needed! Well, I'll be getting all the help while you lot give the help :P
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teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #256 on: January 25, 2013, 10:45:46 am »
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It's a great idea! But I don't think people have time once school starts!
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Re: Chemistry 3/4 2013 Thread
« Reply #257 on: January 25, 2013, 11:48:08 am »
+1
could someone please help me write the balanced half equations for the oxidation and reduction reactions, also when combining the half equations, I get a different answer to the textbook :/
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Re: Chemistry 3/4 2013 Thread
« Reply #258 on: January 25, 2013, 12:50:29 pm »
+1
*2 to balance


This is what I got hope it helps
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Re: Chemistry 3/4 2013 Thread
« Reply #259 on: January 25, 2013, 12:59:16 pm »
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 Is this Heineman chapter 5 Q11d , maybe the textbook answer is wrong
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Re: Chemistry 3/4 2013 Thread
« Reply #260 on: January 25, 2013, 01:17:45 pm »
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Thats the exact same answer I got. hah Yeh maybe the answer is answer is wrong. I'll just wait till the geniuses confirm it. Thanks for your help :)
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Re: Chemistry 3/4 2013 Thread
« Reply #261 on: January 25, 2013, 01:34:59 pm »
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*2 to balance


This is what I got hope it helps
Looks good to me.
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Re: Chemistry 3/4 2013 Thread
« Reply #262 on: January 25, 2013, 02:20:46 pm »
+1
Ok, someone please help me before I kill myself trying to do this question. -.-

For anyone who has the Heinmann Chemistry 2 Book, the question is from Chapter 7, Question 24, pg 106.

The mineral Cobaltite is mined for the production of cobalt. Ore containing cobaltite may also contain trace quantities of nickel. A sample of ore was analysed by AAS to determine the concentration of nickel present. 5.0g of the ore was dissolved in 25ml concentrated nitric acid, then diluted to 100mL. A concentrated stock solution containing 1000ppm nickel was also prepared. 10mL of the ore sample solution was pipetted separately into four 100mL flasks and 1, 2, 4 and 6mL of the concentrated stock nickel solution was added to the flasks. All five flasks were then made up to the mark. A 'standard addition curve was prepared using 10, 20, 40 and 60 ppm added nickel, giving the standards listed in the table.

Standard | Concentration of nickel (ppm) | Absorbance
Standard 1 | x + 10 | 0.25
Standard 2 | x + 20 | 0.34
Standard 4 | x + 40 | 0.44
Standard 6 | x + 60 | 0.53
sample | x  | 0.15

The absorbance readings for the standards and sample were determined using light of wavelength 325.4 nm. The value of x, the concentration of nickel in ppm, was found from the intercept of the calibration curve with the x-asis at absorbance 0.

a) Determine the concentration of nickel in the sample solution by drawing a graph using values listed in the table above. I got the answer which is 16ppm and this is correct.

Now b) What mass of nicel was present in 5.0g of ore?
 
Please and thank you. :)

teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #263 on: January 25, 2013, 02:39:33 pm »
+1
Hey! Thanks for the question.
The important thing to remember in this question is that ppm can be converted into other forms ( ie mg/ ml)

In this case the 16 ppm = 16mg/ml Ni l, right?
Therefore 100ml contains  16 x 100 = 1600 mg from 10 ml of stock = 1.6 mg/l x 100/10
= 16 mg from 5 g of ore.

This q was difficult as! Thanks for asking!
« Last Edit: January 25, 2013, 03:30:20 pm by :) :P :D »
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Re: Chemistry 3/4 2013 Thread
« Reply #264 on: January 25, 2013, 03:09:50 pm »
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If you have any gravimetric analysis questions you're stuck on, post them up here. I have a test/SAC (I'm not sure if it 'counts' or not) on it in the first week back and doing the questions and explaining them back to you would make for some really effective revision. :)
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saba.ay

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Re: Chemistry 3/4 2013 Thread
« Reply #265 on: January 25, 2013, 03:23:28 pm »
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Hey! Thanks for the question.
The important thing to remember in this question is that ppm can be converted into other forms ( ie mg/ ml)

In this case the 16 ppm = 16mg/ml Ni l, right?
Therefore 100ml contains  16 x 100 = 1600 mg from 10 ml of stock = 1.6 mg/l x 100/10
= 1.6mg from 5 g of ore.

This q was difficult as! Thanks for asking!

Damn man, that's what I got too. But apparantly the answer is 16mg. :/ Is my book wrong or am I looking at the wrong answer or something?

teletubbies_95

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Re: Chemistry 3/4 2013 Thread
« Reply #266 on: January 25, 2013, 03:29:16 pm »
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It's 16 mg , I multiplied on my calc wrong it's 1. 6 x 100 /10= 16mg.
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Mr Keshy

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Re: Chemistry 3/4 2013 Thread
« Reply #267 on: January 25, 2013, 03:58:36 pm »
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Totally off topic but excuse me..

I just love how you thank people for asking questions :D, normally I'd expect people to thank each other for answering but you do it both ways :)

Too nice ya!
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Re: Chemistry 3/4 2013 Thread
« Reply #268 on: January 25, 2013, 04:23:02 pm »
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Naww. Thanks. :) and ask more questions!!!
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Mr Keshy

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Re: Chemistry 3/4 2013 Thread
« Reply #269 on: January 25, 2013, 04:28:08 pm »
+1
certainly will :)
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