A 9.02-g sample of liquid water at 40.3 °C is heated by the addition of 203.93 J of energy. The final temperature of the water is °C. The specific heat capacity of liquid water is 4.184 J/gK.

45.70 C

Explanation:

We'll use the formula for specific heat Q=mcΔT. Let's first check out units and look at the values we have.

Q=heat= 203.93 J

m=mass= 9.02g

c= specific heat= 4.184 J/g K

ΔT= Change in temperature= Final temp. - initial temp. = -

Let's plug in what we know.

203.93 J = (9.02g)(4.184)( - 40.3 C)

We'll just solve this algebraically to solve for  . It's like solving for x, just a different variable. It's easier to do the more complicated side first and then start moving your numbers to the other side.

203.93 = (37.73)( - 40.3) We'll multiply 37.73 by   AND 40.3 so we'll get...

203.93 = (37.73Tf - 1520.5)

Let's start moving numbers to the other side

203.93 + 1520.5 = 37.73Tf

1724.4 = 37.73Tf

=

45.70 C=

This makes sense with our problem since they heated water, so we should see an increase in temperature.

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).

explanation: