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The weight of 1*10^22 molecules of CuSo4.5H2O is


User Bandana
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Answer:

To determine the weight of 1*10^22 molecules of CuSO4·5H2O (copper(II) sulfate pentahydrate), we need to calculate the molar mass of the compound and then convert it to grams.

1. Calculate the molar mass of CuSO4·5H2O:

- Copper (Cu) has a molar mass of 63.55 g/mol.

- Sulfur (S) has a molar mass of 32.07 g/mol.

- Oxygen (O) has a molar mass of 16.00 g/mol.

- Hydrogen (H) has a molar mass of 1.01 g/mol.

- Multiply the molar mass of each element by the number of atoms present in the compound and sum them up.

- For CuSO4·5H2O, the molar mass would be:

(63.55 g/mol) + (32.07 g/mol) + (4 × 16.00 g/mol) + (10 × 1.01 g/mol) = 249.68 g/mol.

2. Convert the number of molecules to moles:

- 1*10^22 molecules represents 1 mole of CuSO4·5H2O since Avogadro's number states that 1 mole contains 6.022 * 10^23 molecules.

3. Calculate the weight of 1 mole of CuSO4·5H2O:

- Multiply the molar mass by the number of moles:

249.68 g/mol * 1 mol = 249.68 g.

Therefore, the weight of 1*10^22 molecules of CuSO4·5H2O is 249.68 grams.

Step-by-step explanation:

User Grassyburrito
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