Question

Рассчитайте объем для 160 грамм кислорода о2

Answer 1

Данр:
m (O2) = 160 грамм
_________________
V (O2) = ?

Решение:
V = Vn x n , n = m / M
M (O2) = 32 г/моль
n = 160 г / 32 г/моль = 5 моль
V = 22.4 л x 5 моль = 112 л
Ответ: V (O2) = 112 л

Answer 2

Calculating the Volume of 160 Grams of Oxygen (O2)

To calculate the volume of 160 grams of oxygen (O2), we can use the molar mass of oxygen and the ideal gas law.

Step 1: Calculate the Number of Moles of Oxygen (O2) The molar mass of oxygen (O2) is 32 g/mol. We can calculate the number of moles using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]

\[ \text{moles} = \frac{160 \, \text{g}}{32 \, \text{g/mol}} = 5 \, \text{moles} \]

Step 2: Use the Ideal Gas Law to Calculate the Volume The ideal gas law is given by the formula: \[ PV = nRT \] Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

At standard temperature and pressure (STP), the conditions are: - T = 273 K - P = 1 atm - R = 0.0821 L·atm/(K·mol)

Using STP conditions, we can calculate the volume using the formula: \[ V = \frac{nRT}{P} \]

Substituting the values: \[ V = \frac{(5 \, \text{moles})(0.0821 \, \text{L·atm/(K·mol)})(273 \, \text{K})}{1 \, \text{atm}} \]

\[ V = 11.2 \, \text{liters} \]

Conclusion: The volume of 160 grams of oxygen (O2) at standard temperature and pressure (STP) is 11.2 liters.