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igor_vitrenko [27]
3 years ago
15

What is the number of particles in 27.5 grams of chromium (iii) sulfite ?

Chemistry
1 answer:
aivan3 [116]3 years ago
6 0
<h3><u>Answer;</u></h3>

= 4.81 × 10^22 particles

<h3><u>Explanation</u>;</h3>

The molar mass of Cr2(SO3)3 is 344.18 g

1 mole of a substance contains 6.02 10^23 particles

But; 1 mole of chromium (iii) sulfite = 344.18 g

Therefore;

344.18 g = 6.02×10^23 particles

27.5 g = ?

Hence;

= (27.5 × 6.02 × 10^23)/344.18

<h3><u>= 4.81 × 10^22 particles </u></h3>
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