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October 27, 2025, 03:19:45 pm
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Author Topic: Easy Chem help  (Read 10186 times)  Share 

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andy456

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Easy Chem help
« on: January 14, 2010, 02:40:18 pm »
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Hey,
I just need help with this question. I know that its really simple but its just not working out for me.
Any help is appreciated

Calculate the amount (in mole) of:
 Cl ions in 13.4g of nickel chloride (NiCl2)
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the.watchman

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Re: Easy Chem help
« Reply #1 on: January 14, 2010, 02:45:21 pm »
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Well using with and for NiCl2,



As there are two chloride ions in each molecule of nickel chloride, there is 0.2068mol of Cl ions.

I have only got a headstart on chem and not done any at school, so if I'm wrong, please let me know!
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xD_aQt

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Re: Easy Chem help
« Reply #2 on: January 14, 2010, 02:46:41 pm »
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Quote from: the.watchman on January 14, 2010, 02:45:21 pm
Well using with and for NiCl2,



As there are two chloride ions in each molecule of nickel chloride, there is 0.2068mol of Cl ions.

I have only got a headstart on chem and not done any at school, so if I'm wrong, please let me know!
Haha, damn you beat me to it! :)
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andy456

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Re: Easy Chem help
« Reply #3 on: January 14, 2010, 02:47:27 pm »
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Nah thats right, thanks....
Its like simple revision but I just couldnt get it....
I guess thats what happens when you havent done chemistry in like 3 months
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xD_aQt

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Re: Easy Chem help
« Reply #4 on: January 14, 2010, 02:48:22 pm »
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Quote from: andy456 on January 14, 2010, 02:47:27 pm
Nah thats right, thanks....
Its like simple revision but I just couldnt get it....
I guess thats what happens when you havent done chemistry in like 3 months
You should be alright! Not too late to revise
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andy456

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Re: Easy Chem help
« Reply #5 on: January 14, 2010, 02:49:17 pm »
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Yeah im just going through the first 4 chapters of my text book....
Its pretty much revision off last year.... nothing new yet
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the.watchman

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Re: Easy Chem help
« Reply #6 on: January 14, 2010, 02:50:32 pm »
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Quote from: andy456 on January 14, 2010, 02:49:17 pm
Yeah im just going through the first 4 chapters of my text book....
Its pretty much revision off last year.... nothing new yet


Good luck! I just covered a few chapters of Chem Unit 1
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andy456

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Re: Easy Chem help
« Reply #7 on: January 14, 2010, 02:52:15 pm »
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I hated chem 1..... it took so long for me to understand mole
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xD_aQt

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Re: Easy Chem help
« Reply #8 on: January 14, 2010, 02:56:20 pm »
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Quote from: andy456 on January 14, 2010, 02:52:15 pm
I hated chem 1..... it took so long for me to understand mole
+1 me too ;)
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andy456

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Re: Easy Chem help
« Reply #9 on: January 14, 2010, 03:10:28 pm »
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Thats good to hear.... i thought I was the only one...  ;D
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andy456

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Re: Easy Chem help
« Reply #10 on: January 15, 2010, 12:48:31 am »
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I have another simple question......

Determine the percentage composition of the following compound
Lead (IV) oxide (PbO2)

So I used the empirical formula method to determine:
207.37/239.3688 : 31.9988/239.3688
       0.866         :       0.134
       86.6%        :       13.4%

The answer in the book is 13.4% but in the question it doesn't specify which ion to find the percentage of so i was wondering if the question just means oxygen, or there was like a typo or something in the book.
Btw the book is Heinmann Chemistry Two
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superflya

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Re: Easy Chem help
« Reply #11 on: January 15, 2010, 12:51:29 am »
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oh my bad, i remember doing this question...ur looking for the percentage composition of oxygen.
wording of question is shit.
« Last Edit: January 15, 2010, 01:01:34 am by superflya »
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Potter

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Re: Easy Chem help
« Reply #12 on: January 15, 2010, 01:00:25 am »
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Quote from: andy456 on January 15, 2010, 12:48:31 am
I have another simple question......

Determine the percentage composition of the following compound
Lead (IV) oxide (PbO2)



If the question is written like this, calculate the percentages of all the elements. The book probably just got lazy.
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andy456

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Re: Easy Chem help
« Reply #13 on: January 15, 2010, 01:01:42 am »
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I think I wrote it wrong in the book it says Lead(IV)oxide (PbO2)...

Yeah do textbooks even make typo's....
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superflya

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Re: Easy Chem help
« Reply #14 on: January 15, 2010, 01:03:52 am »
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^^ true and seeing as there are only 2 elements ur pretty much finding both the compositions at the one time.
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