2. HOW MUCH HEAT IS REQUIRED TO BE RELEASED WHEN
COOLING 30,0 GRAMS OF SUPERHEATED WATER VAPOR FROM
120,0°C TO 100.0°C? (SPECIFIC HEAT OF WATER VAPOR IS
187 J/ G°C)

-112200J

Explanation:

The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of water vapour = 30.0g

C = 187 J/ G°C

Φ = (Final temperature - Initial temperature)

= 100°C - 120°C = -20°C

Then apply the formula, Q = MCΦ

Q = 30.0g x 187 J/ G°C x -20°C

Q = -112200J (The negative sign does indicates that heat was released to the surroundings)

Thus, -112200 joules of heat is released when cooling the superheated vapour.

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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the product of the pressure and volume is also constant