A5.00-g sample of liquid water at 25.0 °c is heated by the addition of 84.0 j of energy. the final temperature of the water is °c. the specific heat capacity of liquid water is 4.18 j/g-k.

Final Temperature = 29.2 °C

Explanation:

Before moving any further, let's bring out the information given.

Mass of Liquid water (M) = 5.00g

Initial temperature (T1) = 25 C + 273 = 298 K (Converting to kelvin temperature)

Heat = 84.0 J

FInal Temperature (T2) = ?

Specific Heat Capacity of water (C) =  4.18 J/gK

The formular relating all these parameters is given as;

H = M * C * (T2 -T1)

Making T2 subject of formular, we have;

T2 = (H/MC) + T1

T2 = [84/(5*4.18)] + 298

T2 = 4.02 + 298 = 302 K

Converting to °C, T2 = 302 - 273 = 29.2 °C

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

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