Answer:
B
Explanation:
Recall the law of effusion:
Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.
Let the rate of oxygen be <em>r</em>₁ and the rate of hydrogen be <em>r</em>₂.
The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.
Substitute and solve for <em>r</em>₂:
Because there are 5 moles of hydrogen gas:
In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.
Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.
it is called the partial pressure because it is pressure that is exerted by each gas in a mixture of gases if it occupy the same volume of it own. the the partial pressure of oxygen in the air sample above is 100 mm hg, that of carbon dioxide is 40 mmhg, for water vapor 47 mmhg. and that of nitrogen is 573 mmhg
Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
A. Nonmetallic
B. Nonmetallic
C. Metallic
D. Nonmetallic
E. Metallic
