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ioda
3 years ago
14

How will the solubility of a gas solute change if the pressure above the solution is reduced?

Physics
2 answers:
Hoochie [10]3 years ago
6 0
If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
labwork [276]3 years ago
6 0

Answer:

The limit of the gas's solubility will decrease.

Explanation:

The solubility of a gas in a liquid is explained by Henry's law.

S = k . P

where,

S is the solubility of the gas

k is the Henry's constant (depends on each gas)

P is the partial pressure of the gas (at equilibrium with the pressure above the solution)

We can see that the solubility is directly proportional to the pressure. If the pressure is reduced, the solubility decreases.

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Explanation:

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What is epsilon zero and why does it come into use, particularly in the case of Gauss's Law?
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9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
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<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

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ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

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3 years ago
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8 0
3 years ago
A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o
dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will   its heat output drop if the applied potential difference drops to 110 V ?

we know p = \frac{v^2}{R} .....................(i)

we need to find change in power

\Delta P = \frac{\Delta (V^2)}{R}  

\Delta P = \frac{2 V \Delta V}{R}..............(ii)

from equations we get

\frac{\Delta P}{P} =  \frac{2 \Delta V}{V}

\frac{\Delta P}{P} = 2 \frac{110 -120}{120}

\frac{\Delta P}{P} =  -2(\frac{10}{120})

\frac{\Delta P}{P} = - 0.16 %

(b)

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and due to this increment of heating power resistance will decrease so actual drop in the power would  be smaller

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