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kipiarov [429]
3 years ago
8

Help me the question number 1 , please !

Chemistry
2 answers:
Igoryamba3 years ago
7 0
The answer is 10x3.01x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x3.e+22=9.06e+44
maksim [4K]3 years ago
4 0
10•3.01•210•3.e+22=9.06e+44
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What mass of iron (III) nitrate will be in 129.8ml of a 0.3556 molar aq iron (III) nitrate solution?
Anvisha [2.4K]

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\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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0.3556M=\frac{\text{Mass of iron (III) nitrate}\times 1000}{241.86 g/mol\times 129.8}\\\\\text{Mass of iron (III) nitrate}=11.16g

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