Where are the answers that apply .. hopes this help :)
Answer:
14175 j heat released.
Explanation:
Given data:
Mass of aluminium = 350.0 g
Initial temperature = 70.0°C
Final temperature = 25.0°C
Specific heat capacity of Aluminium = 0.9 j/g.°C
Heat changed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
Heat change:
ΔT = Final temperature - initial temperature
ΔT = 25.0°C - 70°C
ΔT = -45°C
Q = m.c. ΔT
Q = 350 g × 0.9 j/g.°C × -45°C
Q = -14175 j
Answer: 64
Explanation:
you just multiple the 3 numbers to get the answer (i’m in chemistry and just did this question lol)
Answer:
pV= nRT
Explanation:
(p1 × V1)/ T1/ (p2 × V2)/ T2
