Stoichiometry moles helpp
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Stoichiometry moles helpp

[From: ] [author: ] [Date: 11-08-19] [Hit: ]
1.1416mg Mg(NO3)2 or 1.The molar mass of Mg(NO3)2 is 24.31+2(14.01)+6(16)=148.Hope you can read all that.......
Now if your lab partner determined that there were 3.448 x 10(to the22 )atoms of O in an unknown sample of Mg(NO3)2, how many milligrams would the unknown sample weigh?

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[3.448*10^22 atoms 0] * [1mol O/(6.02*10^23 atoms O)] * [1 mol Mg(NO3)2/6 mol O] *
[148.33g Mg(NO3)2/1mol Mg(NO3)2]

1.416g Mg(NO3)2 * 1000 =



1416mg Mg(NO3)2 or 1.416*10^3 mg Mg(NO3)2



The molar mass of Mg(NO3)2 is 24.31+2(14.01)+6(16)=148.33g Mg(NO3)2

Hope you can read all that.
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