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How much would 3.01x10^23 atoms of silver nitrate weigh?

Sagot :

Explanation:

To solve this question, we need to know the Avogadro's constant:

6.022 x 10^23 atoms equal 1 mol.

So, in this case we have 3.01x10^23

We first transform it to moles and then to grams, using the following formula:

m = n*MM

where:

m = mass

n = moles

MM = molar mass of silver nitrate (AgNO3) = 169.87 g/mol

6.022 x 10^23 ---- 1 mol

3.01x10^23 ---- x mol

x = 0.5 moles

m = 0.5 * 169.87 g/mol

m = 85 g

Answer: 85 g

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