A. when the volume of the gas is halved, the pressure of the gas will be doubled.
<h3>Pressure of the gas according to ideal gas law</h3>
PV = nRT
P = nRT/V
when the volume is halved
P= nRT/0.5V
P = 2nRT/V
P(new) = 2P(initial)
Thus, when the volume of the gas is halved, the pressure of the gas will be doubled.
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Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
So the number of moles = 0.06 moles of 4X12-2 moles.
The equilibrium constant expression for this reaction that takes place in water and involve ions can be written as K= [H_3 O^+ ][OH^- ]/([H_2 O] ^2 ). But the concentration of undissociated water, H_2 O is much larger than the concentration of the ions that is essentially remains constant. Therefore, we can include it in the equilibrium constant. The resulting new equilibrium constant can be written: K_W= [H_3 O^+ ][OH^- ].
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