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3 years ago

A flask at room temperature contains exactly equal amounts (moles of nitrogen and oxygen.

1 answer:

7 0

Equal amounts (moles of nitrogen and oxygen).

a. which gas has the higher partial pressure?

Partial pressure is proportional to the molar fraction of the gas.

Given that there are equal amounts, the molar fractions and the partial pressures are equal.

b. molecules of which gas have the higher average velocity?

The average velocity is inversely related to the square root of molar mass of the gas, then the gass with the lower molar mass will have the higher average velocity.

Molar mass of N2 = 2 * 14 g/mol = 28 g/mol
Molar mass of O2 = 2 * 16 g/mol = 32 g/mol

Then, N2 will have a higher average velocity.

c. the molecules of which gas effuses more quickly if a small hole was opened in the flask?

The effusion rate is inversely related to the square root of the molar mass, then the gas with the lower molar mass will effuse more quickly.

Therefore, N2 will effuse more quickly.

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Why do molecules combined into chains?

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3 years ago

The following equation represents the type of reaction called

Mazyrski [523]

Answer:

Reduction

Explanation:

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

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Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.

Examples:

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2 years ago

You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

bonufazy [111]

Answer:

$\frac{[F^{-}]}{[HF]}$ is larger

Explanation:

$pK_{a}=-logK_{a}$ , where $K_{a}$ is the acid dissociation constant.

For a monoprotic acid e.g. HA, $K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$ and $\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}$

So, clearly, higher the $K_{a}$ value , lower will the the $pK_{a}$

In this mixture, at equilibrium, $[H^{+}]$ will be constant.

$K_{a}$ of HF is grater than $K_{a}$ of HCN

Hence, $(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})$

So, $\frac{[F^{-}]}{[HF]}$ is larger

5 0

3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

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3 years ago

How can you find the charge on po4

Anvisha [2.4K]

Answer:

Answer:

Explanation:

I hope it's helpful!

7 0

3 years ago

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