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At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of

Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of carbon dioxide that can be dissolved is 1.7 grams

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

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How does air pressure affect the boiling point of a liquid?

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Answer:

D. Air pressure lowers the temperature of the liquid molecules.

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If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

$\tt \dfrac{8.5}{78.11}=0.109$

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

$\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol$

Heat transferred to water :

Q=m.c.ΔT

$\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC$

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Which statements about salt water are true? Check all that apply.

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Salt water contains minerals other than salt.

The temperature of ocean water is affected by the depth of the water.

The temperature of ocean water is affected by the ocean’s location.

Objects float more easily in oceans than in lakes or rivers.

These are all of the true answers

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Number of hydrogen atom(s) in water (H2O)

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Answer:

Maybe

Explanation:

the number of hydrogen atom in water is 2..

hope it helps

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