3 years ago

1.03 – 0.06 subtracting Desimals of Different Lengths 1c

Mathematics

1 answer:

kkurt [141]3 years ago

6 0

The answer should be 0.97

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But 1 mole is equivalent to its molar mass in terms of weight, therefore;

${ \sf{6.02 \times {10}^{23} \: molecules = 44 \: g }}\\ { \sf{8.11 \times {10}^{26} \: molecules =( \frac{8.11 \times {10}^{26} }{6.02 \times {10}^{23} } \times 44) \: g}} \\ \\ = { \boxed{ \boxed{ \bf{59275.7 \: \: g}}}}$

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