The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at 
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
Answer:
Explanation:
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In this case, according to the reaction:
Thus, since there is a 4:2 mole ratio of P to P2Cl5 and the molar mass of the later is 239.2125 g/mol, we obtain the following mass as the produced one:
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Answer:
1.53 L
Explanation:
Step 1: Given data
- Mass of oxygen (m): 11.2 g
- Ideal gas constant (R): 0.0821 atm.L/mol.K
Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
11.2 g × (1 mol/32.00 g) = 0.350 mol
Step 3: Calculate the volume of oxygen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm
V = 1.53 L
Answer:
Hydroxy(oxo)bismuthine oxide
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