To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g;
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348
= 26250.0348 J or 26.250 kJ
Answer:
C3 H6 O2
Explanation:
first divide their mass by their respective molar mass, we get:
30.4 moles of C
61.2 moles of H
20.25 moles of O
now divide everyone by the smallest one of them then we get
C= 1.5
H= 3
O= 1
since our answer of C is not near to any whole number so we will multiply all of them by 2
so,
C3 H6 O2 is our answer
Answer:
change in the total mass of substances
