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How many nitrate ions are present in 4.18 x 10-20 grams of copper (II) nitrate?

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Answer:

4.46 x 10^-22 nitrate ions

Step-by-step explanation:

To find the number of nitrate ions present in 4.18 x 10^-20 grams of copper (II) nitrate, we need to use the molar mass of copper (II) nitrate and Avogadro's number.

The molar mass of copper (II) nitrate is:

Cu(NO3)2 = 63.55 g/mol (Cu) + 2(14.01 g/mol (N) + 3(16.00 g/mol (O))) = 187.57 g/mol

So 1 mole of copper (II) nitrate has a mass of 187.57 grams.

We can find the number of moles of copper (II) nitrate in 4.18 x 10^-20 grams by dividing the mass by the molar mass:

4.18 x 10^-20 g / 187.57 g/mol = 2.23 x 10^-22 moles

Each mole of copper (II) nitrate contains 2 moles of nitrate ions. Therefore, the total number of nitrate ions in 4.18 x 10^-20 grams of copper (II) nitrate is:

2.23 x 10^-22 moles x 2 nitrate ions/mole = 4.46 x 10^-22 nitrate ions

So there are approximately 4.46 x 10^-22 nitrate ions in 4.18 x 10^-20 grams of copper (II) nitrate.

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