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Kobotan [32]
3 years ago
14

he solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at

2.5 atm and 25 °C is ________ g/L.
Chemistry
1 answer:
Tema [17]3 years ago
8 0

Answer:

0.1025 g/L

Explanation:

From the question above,

The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L

Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure

Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows

= 0.041/1.0 × 2.5

= 0.041×2.5

= 0.1025 g/L

Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L

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Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr
vredina [299]

Answer : The correct expression for equilibrium constant will be:

K_c=\frac{[C]^8}{[A]^4[B]^2}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

4A+2B\rightleftharpoons 8C

The expression of K_c will be,

K_c=\frac{[C]^8}{[A]^4[B]^2}

Therefore, the correct expression for equilibrium constant will be, K_c=\frac{[C]^8}{[A]^4[B]^2}

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3 years ago
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What is the balance equation for HgO(s)- Hg(l)+O2(g)
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You are practicing classification by sorting pictures into groups based on the characteristics of the kingdoms. While you are so
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If you are given a 0. 31 g piece of sodium metal to react with water, how many moles of hcl would it take to neutralize the sodi
Sphinxa [80]

The moles of HCl to neutralize the sodium hydroxide produced is<u> 0.0135 mole.  </u>

Neutralization or neutralization is a chemical response wherein an acid and a base react quantitatively with each other. In a reaction in water, neutralization outcomes in there being no excess of hydrogen or hydroxide ions gift in the answer.

<u>calculation:-</u>

<u />

2Na + 2H₂O  -----> 2NaOH + H₂

2 mol or 46g of Na produces 80 grams of NaOH

∴ 0.31 g of Na will produce = 80/46 × 0.31

                                              =  0.54 gram of NaOH.

mol of NaOH = 0.54/40

                      = 0.0135

Since both Hcl and NaOH have the same valance factor,

1 mole NaOH is needed to neutralize 1 mol HCl

∴ 0.0135 mole of NaOH will need = 0.0135 mole of HCl

mass = 0.0135 × 36.5

         =<u> 0.493 grams of HCL.</u>

Learn more about neutralizing here:-brainly.com/question/23008798

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