Answers:
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This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
Answer:
-191.7°C
Explanation:
P . V = n . R . T
That's the Ideal Gases Law. It can be useful to solve the question.
We replace data:
2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°
(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°
T° = 81.3 K
We convert T° from K to C°
81.3K - 273 = -191.7°C
Answer:
Bohr model A
Explanation:
It has more valance electrons therefore has more interaction between the atoms and has more electronegativity.
