To solve the problem, we can use Avogadro's number, which is 6.022 × 10²³. This number represents the number of particles (atoms, molecules, ions) in one mole of a substance.
First, we need to calculate the number of potassium atoms in one mole:
1 mol K = 6.022 × 10²³ atoms K
Then, we can use this conversion factor to calculate the number of potassium atoms in 3.227 × 10¹⁰ mol of potassium:
3.227 × 10¹⁰ mol K × 6.022 × 10²³ atoms K/mol K = 1.942 × 10³⁴ atoms K
Therefore, there are 1.942 × 10³⁴ atoms of potassium in 3.227 × 10¹⁰ mol of potassium.
Calculate the number of atoms in 3.227*10 ^10mol of potassium
3 answers
Find mass of. H atomms. Present in 32g Methane. CH4.
To find the mass of H atoms in 32 g of methane, we first need to calculate the number of moles of methane present in 32 g. The molar mass of methane (CH4) is 16 g/mol (12 g/mol for carbon and 4 g/mol for hydrogen).
Number of moles of CH4 = 32 g / 16 g/mol = 2 mol
Next, we can use the balanced chemical equation for the combustion of methane to determine the ratio of moles of H atoms to moles of CH4:
CH4 + 2O2 → CO2 + 2H2O
For every mole of methane, there are 2 moles of H2O produced, which contains 4 moles of H atoms.
So, for 2 moles of CH4, there are 4 moles of H2O and 8 moles of H atoms.
Finally, we can calculate the mass of 8 moles of H atoms:
Mass of 8 moles of H atoms = 8 mol x 1.008 g/mol (molar mass of H) = 8.064 g
Therefore, there are 8.064 g of H atoms present in 32 g of methane.
Number of moles of CH4 = 32 g / 16 g/mol = 2 mol
Next, we can use the balanced chemical equation for the combustion of methane to determine the ratio of moles of H atoms to moles of CH4:
CH4 + 2O2 → CO2 + 2H2O
For every mole of methane, there are 2 moles of H2O produced, which contains 4 moles of H atoms.
So, for 2 moles of CH4, there are 4 moles of H2O and 8 moles of H atoms.
Finally, we can calculate the mass of 8 moles of H atoms:
Mass of 8 moles of H atoms = 8 mol x 1.008 g/mol (molar mass of H) = 8.064 g
Therefore, there are 8.064 g of H atoms present in 32 g of methane.