A100.0 g sample of water at 27.0°c is poured into a 70.3 g sample of water at 89.0°c. what will be the final temperature of the water?

The final temperature of water = 52.6 C

Explanation:

The heat (q) lost or gained by a substance of mass m corresponding to a temperature change from T1 to T2 degrees is given as:

-----(1)

where c = specific heat of the substance

For water, c = 4.18 J/gC

In the given situation:

Heat lost by 70.3 g of water = Heat gained by 100.0 g of water

Based on equation (1) and considering that heat lost is negative:

Solving the above equation gives:

T2 = 52.6 C

the answer would be b i believe

the ratio of the conjugate base to the acid required to attain the desired ph.