A tank contains 7.0 moles of a mixture of nitrogen gas and oxygen gas. The total

Chemistry

1 answer:

Pavel [41]2 years ago

8 0

Answer:

5.2 mol

Explanation:

Step 1: Given data

Total number of moles (n): 7.0 mol

Total pressure (P): 530 kPa

Partial pressure of oxygen (pO₂): 140 kPa

Step 2: Calculate the mole fraction of oxygen

We will use the following expression.

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 140 kPa / 530 kPa

X(O₂) = 0.264

Step 3: Calculate the moles of oxygen

We will use the definition of mole fraction of oxygen.

X(O₂) = n(O₂) / n

n(O₂) = X(O₂) × n

n(O₂) = 0.264 × 7.0 mol

n(O₂) = 1.8 mol

Step 4: Calculate the moles of nitrogen

The total number of moles is equal to the sum of moles of the individual gases.

n(O₂) + n(N₂) = n

n(N₂) = n - n(O₂)

n(N₂) = 7.0 mol - 1.8 mol

n(N₂) = 5.2 mol

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Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.