Can someone please help me! This is due tomorrow and I have no idea how to do it

1.54 ml

2. 54 g

3. -43.7 °C

4. 97 g

5. 17.74 kJ

Explanation:

1. Volume of water that only came from the melted ice is;

Volume of water plus melted ice-volume of warm water

151 ml - 97 ml = 54 ml

2.Mass of water that came from the melted ice.

Mass=density of water * volume of water from melted ice

M=1 g/ml * 54 ml = 54 g

3. The change in temperature of warm water will be;

Lowest temperature of ice-water mixture-Temperature of warm water

ΔT=0.9-44.6 = -43.7

4. Mass of warm water will be;

Mass of warm water= density of water*volume of warm water

M=1g/ml * 97 = 97 g =0.097 kg

5.

Energy released by warm water as it cools

Q=mcΔT

Given , mass of warm water=m=97

c=4.184 j/g*°C = 4184 J/(kg.C)

Δ= - 43.7

Q=0.097*4184*43.7 =17735.56 J

17735.56/1000 =17.74 kJ

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