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2 years ago

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A container contains a mixture of hydrogen, oxygen and nitrogen gases. The pressure of hydrogen is 78 kPa. The pressure of oxyge

n is 2.4416kPa. The pressure of nitrogen is 59.8 kPa. What is the total pressure of the container?

1 answer:

Nuetrik [128]2 years ago

7 0

<u>Answer:</u> The total pressure of the container is 140.2416 kPa

<u>Explanation:</u>

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the total pressure, we use the law given by Dalton, which is:

$P_T=p_{H_2}+p_{O_2}+p_{N_2}$

We are given:

Vapor pressure of hydrogen gas, $p_{H_2}$ = 78 kPa

Vapor pressure of oxygen gas, $p_{O_2}$ = 2.4416 kPa

Vapor pressure of nitrogen gas, $p_{N_2}$ = 59.8 kPa

Putting values in above equation, we get:

$P_T=78+2.4416+59.8\\\\P_T=140.2416kPa$

Hence, the total pressure of the container is 140.2416 kPa

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Answer:

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Explanation:

To solve the question

The molarity ratio is given by

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To find the mass, we note that the Number of moles = $\frac{Mass}{Molar Mass}$ from which we have

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The Lewis-dot structure of perbromate ion is shown below.

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Read 2 more answers

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

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Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

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