1. To determine the number of moles in 25.0g of O₂, we need to use the molar mass of O₂. The molar mass of O₂ is approximately 32.00 g/mol. We can use the formula:
moles = mass (g) / molar mass (g/mol)
Plugging in the values, we get:
moles = 25.0g / 32.00 g/mol ≈ 0.78125 mol
Therefore, there are approximately 0.78125 moles in 25.0g of O₂.
2. To determine the number of particles in 100.0g of gold, we need to use Avogadro's number, which is approximately 6.022 x 10²³ particles/mol. The molar mass of gold (Au) is approximately 196.97 g/mol. We can use the formula:
particles = moles * Avogadro's number
First, let's find the number of moles:
moles = mass (g) / molar mass (g/mol) = 100.0g / 196.97 g/mol ≈ 0.5072 mol
Now, we can calculate the number of particles:
particles = 0.5072 mol * 6.022 x 10²³ particles/mol ≈ 3.057 x 10²³ particles
Therefore, there are approximately 3.057 x 10²³ particles in 100.0g of gold.
3. To determine the number of liters in 3 moles of carbon dioxide (CO₂), we need to use the ideal gas law and assume standard temperature and pressure conditions (STP). At STP, 1 mole of any gas occupies 22.4 liters.
liters = moles * 22.4 L/mol = 3 mol * 22.4 L/mol = 67.2 L
Therefore, there are 67.2 liters in 3 moles of carbon dioxide at STP.
4. To determine the number of particles in 50.0L of NO₂, we need to use Avogadro's number. Since NO₂ is a gas, we assume ideal gas behavior.
particles = (volume (L) * Avogadro's number) / molar volume
At STP, the molar volume of any gas is 22.4 L/mol. Therefore:
particles = (50.0 L * 6.022 x 10²³ particles/mol) / 22.4 L/mol ≈ 1.35 x 10²³ particles
Therefore, there are approximately 1.35 x 10²³ particles in 50.0L of NO₂ at STP.
5. To determine the number of grams in 5.0 x 10²⁶ molecules of water (H₂O), we need to use the molar mass of water. The molar mass of water is approximately 18.02 g/mol.
grams = (number of molecules * molar mass) / Avogadro's number
Plugging in the values, we get:
grams = (5.0 x 10²⁶ molecules * 18.02 g/mol) / (6.022 x 10²³ particles/mol) ≈ 1.51 x 10³ g
Therefore, there are approximately 1.51 x 10³ grams in 5.0 x 10²⁶ molecules of water.
6. To determine the number of grams in 6.00 moles of C₆H₁₂O₆, we need to use the molar mass of C₆H₁₂
O₆. The molar mass of C₆H₁₂O₆ is approximately 180.18 g/mol.
grams = moles * molar mass
Plugging in the values, we get:
grams = 6.00 mol * 180.18 g/mol = 1081.08 g
Therefore, there are 1081.08 grams in 6.00 moles of C₆H₁₂O₆.
7. To determine the number of moles in 22.4L of H₂S, we need to use the ideal gas law and assume standard temperature and pressure conditions (STP). At STP, 22.4L of any gas is equal to 1 mole.
moles = volume (L) / molar volume
moles = 22.4 L / 22.4 L/mol = 1 mol
Therefore, there is 1 mole in 22.4L of H₂S at STP.
8. To determine the number of moles in 3.00 x 10²² atoms of gold, we need to use Avogadro's number. Since 1 mole of any substance contains 6.022 x 10²³ particles (atoms, molecules, etc.), we can use the following conversion:
moles = number of particles / Avogadro's number
Plugging in the values, we get:
moles = 3.00 x 10²² atoms / (6.022 x 10²³ atoms/mol) ≈ 4.98 x 10⁻² moles
Therefore, there are approximately 4.98 x 10⁻² moles in 3.00 x 10²² atoms of gold.
9. To determine the number of particles in 25.0g of carbon (C), we need to use the molar mass of carbon and Avogadro's number.
moles = mass (g) / molar mass (g/mol)
moles = 25.0g / 12.01 g/mol ≈ 2.08 mol
particles = moles * Avogadro's number
particles = 2.08 mol * 6.022 x 10²³ particles/mol ≈ 1.25 x 10²⁴ particles
Therefore, there are approximately 1.25 x 10²⁴ particles in 25.0g of carbon.
10. To determine the number of liters in 20.0 moles of carbon dioxide (CO₂) at STP, we can use the molar volume of a gas at STP, which is 22.4 L/mol.
liters = moles * molar volume
liters = 20.0 mol * 22.4 L/mol = 448 L
Therefore, there are 448 liters in 20.0 moles of carbon dioxide at STP.
11. To determine the number of grams in 44.8L of oxygen (O₂) at STP, we need to use the molar mass of oxygen, which is approximately 32.00 g/mol.
moles = volume (L) / molar volume
moles = 44.8 L / 22.4 L/mol = 2 mol
grams = moles * molar mass
grams = 2 mol * 32.00 g/mol = 64.00 g
Therefore, there are 64.00 grams in 44.8L of oxygen at STP.
12. To determine the number of liters in 1.34g of carbon monoxide (CO), we need to use the molar mass of carbon monoxide,
which is approximately 28.01 g/mol.
moles = mass (g) / molar mass (g/mol)
moles = 1.34g / 28.01 g/mol ≈ 0.0478 mol
liters = moles * molar volume
liters = 0.0478 mol * 22.4 L/mol ≈ 1.07 L
Therefore, there are approximately 1.07 liters in 1.34g of carbon monoxide.
