Question

How many atoms of Cl2 gas are in 38.7L?
What is the mass of 2.3 x 10^24 particles of Aluminum Sulfide?
If I have 425 grams of water, how many particles of water do I have?
How many liters of space is occupied by 75.0 g of CO2 gas?

Answer 1

Answer on Question#48455 – Chemistry – Other

How many atoms of Cl2 gas are in 38.7L?

What is the mass of $2.3 \times 10^{24}$ particles of Aluminum Sulfide?

If I have 425 grams of water, how many particles of water do I have?

How many liters of space is occupied by $75.0\ \mathrm{g}$ of CO2 gas?

Solution:

$$v = \frac{m}{M}; \quad v - \text{mole (mol)}; \quad m - \text{the mass (g)}; \quad M - \text{the molar mass (g mol}^{-1}).$$ $$v = \frac{N}{N_{A}}; \quad N - \text{the number of particles (atoms, molecules)}; \quad N_{A} = 6.02 \times 10^{23} \ \mathrm{mol}^{-1} \quad \text{(Avogadro constant)}$$ $$v = \frac{V}{V_{m}}; \quad V - \text{the volume (L)}; \quad V_{m} \text{ (the molar volume)} = 22.4 \ \mathrm{L mol}^{-1}$$

How many atoms of Cl2 gas are in 38.7L?

$$N(\mathrm{Cl2}) = 1.04 \times 10^{24} \ \mathrm{N}(\mathrm{Cl}) = 2.08 \times 10^{24}$$

Answer: $N(\mathrm{Cl}) = 2.08 \times 10^{24}$

What is the mass of $2.3 \times 10^{24}$ particles of Aluminum Sulfide?

$$m(\mathrm{Al2S3}) = 573\ \mathrm{g}$$

Answer: $573\ \mathrm{g}$

If I have 425 grams of water, how many particles of water do I have?

$$v = \frac{m}{M}; \quad v = \frac{N}{N_{A}}; \quad N(\mathrm{H2O}) = 1.42 \times 10^{25}$$

Answer: $1.42 \times 10^{25}$

How many liters of space is occupied by $75.0\ \mathrm{g}$ of CO2 gas?

$$v = \frac{m}{M}; \quad v = \frac{V}{V_{m}}; \quad V(\mathrm{CO2}) = 38.2\ \mathrm{L}$$

Answer: $V(\mathrm{CO2}) = 38.2\ \mathrm{L}$

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