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makkiz [27]
4 years ago
8

How many moles are in 9.571x10^2 mega grams of nickel (iii) sulfate?

Chemistry
1 answer:
Leni [432]4 years ago
8 0

First convert mega grams to grams

(9.57x10^2)*(10^6)=957100000 grams

Then find the molar mass of nickel sulphate

Nickel Ni 3+ and sulphate SO4 2+

So formula= Ni2(SO4)3

So mr = (59*2) + (32*3) + (16*12)

= 406gmol-1

Moles = mass/mr

957100000/406

=2357389.163 moles


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