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Yuki888 [10]
3 years ago
6

The solubility of O2 in water is 0.590 g/L at an oxygen pressure of around 15.5 atm. What is the Henry's law constant for O2?

Chemistry
1 answer:
Lana71 [14]3 years ago
8 0

<u>Answer:</u> The Henry's constant for oxygen gas is 3.81\times 10^{-2}g/L.atm

<u>Explanation:</u>

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

C_{O_2}=K_H\times p_{O_2}

where,

K_H = Henry's constant = ?

C_{O_2} = molar solubility of oxygen gas = 0.590g/L

p_{O_2} = partial pressure of oxygen gas = 15.5 atm

Putting values in above equation, we get:

0.590g/L=K_H\times 15.5atm\\\\K_H=\frac{0.590g/L}{15.5atm}=3.81\times 10^{-2}g/L.atm

Hence, the Henry's constant for oxygen gas is 3.81\times 10^{-2}g/L.atm

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Answer:

Heat flows from the block at high temperature to the one with lower temperature

Explanation:

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7 0
3 years ago
assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the in
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Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

<em>Based on the equilibrium:</em>

<em>Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺</em>

<em>kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]</em>

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

<em> </em>

<em> </em>

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7 0
2 years ago
I REALLY NEED HELP ITS DUE AT 11:59 PM
Slav-nsk [51]

Answer:

I attached a photo of balanced equations but thats as much as I can help.

Explanation:

5 0
3 years ago
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Explanation:

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Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution&#10;}}

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Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

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nataly862011 [7]

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1.27*10^-36=x^2

Take the square root of both sides.

1.13*10^-18=x

That would be your solubility.

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3 years ago
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