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Topic: Aluminium liberation H2  (Read 9855 times)

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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #15 on: November 15, 2009, 05:08:09 PM »
Thank you for your help
goodnight
 :)

Offline huskywolf

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Re: Aluminium liberation H2
« Reply #16 on: November 15, 2009, 07:04:42 PM »
Actually 1 more question please. 5g Ca added to  HCl

using the weight of the metal used and moles of H2gas produced calculate the atomic weight of the metal.
so for Ca
n=V/Vm, n(h2)=0.232L/24.1Lmol so 0.010 mol (H2)

M=n/m ,M(Ca)=0.010mol/0.5g=0.02gmol?             {M=atomic weight}

0.02gmol could not be the atomic weight of calcium in this reaction could it?
« Last Edit: November 15, 2009, 07:41:29 PM by huskywolf »

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Re: Aluminium liberation H2
« Reply #17 on: November 16, 2009, 07:47:12 AM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline huskywolf

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Re: Aluminium liberation H2
« Reply #18 on: November 16, 2009, 10:29:56 AM »
Our tutor gaves us that...I thought he was wrong
Could you tell me what the correct formula is to calculate atomic weight from lab data gathered Please

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