Answer:
Q = 1461.6 J
Explanation:
Given data:
Mass of ice = 36 g
Initial temperature = -20°C
Final temperature = 0°C
Amount of heat absorbed = ?
Solution:
specific heat capacity of ice is 2.03 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 0°C - (-20°C)
ΔT = 20°C
Q = 36 g ×2.03 j/g.°C×20°C
Q = 1461.6 J
The answer is statement #3.
Solution:
mass of the cellulose in the mixture is 0.38g
total mass of the mixture is:
3.35+0.38+8.76
=12.4g
thus the percentage of the cellulose in the mixture is:
mass of the cellulose/total mass of the mixture*100%
0.38/12.4*100%
=3%
Answer:17.9 grams
Explanation:I took the quiz
Hope this helps
