Answer: The partial pressure of oxygen is 187 torr.
Explanation:
According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.
where, x = mole fraction
= total pressure
,
,
Thus the partial pressure of oxygen is 187 torr.
This
reaction is called the electrolysis of water. The balanced reaction is:
2H2O = 2H2 + O2
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We are given the amount of water for the electrolysis reaction. This
will be the starting point of our calculation.
45.6 grams H2O (1 mol H2O / 18.02 g H2O) (1 mol O2 / 2 mol H2O) = 1.27 mol O2
V = nRT/P = </span><span>1.27 mol O2 (0.08206 atm L / mol K) (301 K) / 1.24 atm
V = 25.20 L O2</span>
Answer:
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized
O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol
The answer is A. you sre correct!
