Question

1.A flask contains 0.230 mol of liquid bromine, Br2. Determine the number of bromine molecules present in the flask.

Express your answer numerically in molecules.
2.Calculate the mass of 1.00×1024 (a septillion) molecules of water.
Express the answer numerically in grams.
3.The fuel used in many disposable lighters is liquid butane, C4H10. How many carbon atoms are in 3.00 gof butane?
Express your answer numerically in atoms

Answer 1

The number of bromine molecules in 0.230 mol of Br2 is 1.385 × 1023. The mass of 1.00×1024 water molecules is 29.89 g. There are 1.238 × 1023 carbon atoms in 3.00 g of butane.

Step-by-step explanation:

To determine the number of bromine molecules in 0.230 mol of liquid bromine (Br2), we use Avogadro's number, which states that 1 mol contains 6.022 × 1023 molecules. To calculate mass and number of atoms, we also need the molar masses from the periodic table.

1. To find the number of bromine molecules in 0.230 mol of Br2:
2. Number of molecules = moles × Avogadro's number = 0.230 mol × 6.022 × 1023 molecules/mol.
3. Number of molecules = 1.385 × 1023 molecules of Br2.
4. To calculate the mass of 1.00×1024 molecules of water (H2O):
5. The molar mass of water is approximately 18.01 g/mol.
6. Mass = (number of molecules / Avogadro's number) × molar mass = (1.00×1024 / 6.022 × 1023) × 18.01 g/mol.
7. Mass = 29.89 g of water.
8. For the number of carbon atoms in 3.00 g of butane (C4H10):
9. Molar mass of butane is approximately 58.12 g/mol.
10. Moles of butane = mass / molar mass = 3.00 g / 58.12 g/mol.
11. Moles of butane = 0.0516 mol.
12. Number of carbons = moles of butane × 4 (since C4H10) × Avogadro's number.
13. Number of carbon atoms = 0.0516 mol × 4 × 6.022 × 1023 atoms/mol.
14. Number of carbon atoms = 1.238 × 1023 atoms of carbon.