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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.[tex][/tex]
