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anzhelika [568]

3 years ago

10

a sample of hydrogen gas (h2) is mixed with water vapor (h2o (g)). the make sure has a total pressure of 811 torr, and the water

vapor has a partial pressure of 12 torr. how many moles of hydrogen gas are present in a 10.0 l mixture at 298k?

1 answer:

Alborosie3 years ago

6 0

Answer:

$n=0.430molH_2$

Explanation:

Hello!

In this case, considering the partial Dalton's law of partial pressures, we can notice that the total pressure equals the pressure of steam and the pressure of hydrogen, which can be determined as shown below:

$p_T=p_H+p_w\\\\p_H=811torr-12torr=799torr*\frac{1atm}{760torr}\\\\p_H=1.05atm$

Thus, by using the ideal gas law, we can compute the moles of hydrogen as shown below:

$PV=nRT\\\\n= \frac{PV}{RT}=\frac{1.05atm*10.0L}{0.082\frac{atm*L}{mol*K}*298K}\\\\n=0.430molH_2$

Best regards!

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2 years ago

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .

3 0

3 years ago

1. A chemical equation is balanced when *

lubasha [3.4K]

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6 0

3 years ago

Read 2 more answers

Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a

horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

Ni 12.7 g

CO 1.98 g

$Ni(CO)_4$ 0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of $Ni(CO)_4$ = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

$\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M$

$\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M$

$\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

The expression of equilibrium constant for the reaction:

$K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}$

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1$

Hence, the value of equilibrium constant for the reaction is 2448.1

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