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#Post#: 2393--------------------------------------------------
I need some help
DIR By: suzidemello
Date: March 29, 2013, 5:33 am
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I need some help. I have a question but i am not sure about its
solution.
Calculate the number of ammonium ions present in 1.32g of
ammonium sulphate ?
Please help.
#Post#: 2394--------------------------------------------------
Re: I need some help
DIR By: Shiva
Date: March 29, 2013, 5:43 am
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0.36g as far as I know.
Shiva
#Post#: 2397--------------------------------------------------
Re: I need some help
DIR By: Michel
Date: March 29, 2013, 6:12 am
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--- Quote from: Shiva link ---
>
> 0.36g as far as I know.
>
> Shiva
>
--- End Quote ---
number, not mass ;)
#Post#: 2399--------------------------------------------------
Re: I need some help
DIR By: Michel
Date: March 29, 2013, 6:25 am
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--- Quote from: suzidemello link ---
>
> I need some help. I have a question but i am not sure about
its solution.
>
> Calculate the number of ammonium ions present in 1.32g of
ammonium sulphate ?
>
> Please help.
>
>
--- End Quote ---
Show your attempt in solving the equation. where did you become
unsure of the solution,? then we can help
#Post#: 2456--------------------------------------------------
Re: I need some help
DIR By: suzidemello
Date: April 11, 2013, 4:19 am
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It will be 1.32g (NH4)2SO4 x (1mol)/(molar mass of (NH4)2SO4) x
2mol/1mol = mol of NH4+
I found this on yahoo answers.
Can you help with the exact answer.
#Post#: 2458--------------------------------------------------
Re: I need some help
DIR By: Michel
Date: April 11, 2013, 7:02 am
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Write the equation for ionization of ammonium sulfate. Then
notice one mole of ammonium sulfate will produce 2 moles of
NH4+, multiply the number of moles of NH4+ you calculated by 2
to get the number of moles of NH4+, After which you multiply by
Avogadro constant to get the number of NH4+
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