63 mm multiply the length value by 10 bb ♡
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<u>Given:</u>
Mass of Ba = 1.50 g
Mass of H2O = 100.0 g
Initial temp T1 = 22 C
Final Temp T2 = 33.1 C
specific heat c = 4.18 J/g c
<u>To determine:</u>
The reaction enthalpy
<u>Explanation:</u>
The heat released during the reaction is:
q = - mc(T2-T1) = - (100+1.5) g *4.18 J/g C * (33.1-22) C = -4709.4 J
# moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g.mol-1 = 0.0109 moles
ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol
Ans : The enthalpy change for the reaction is -432 kJ/mol
Answer:
13598 J
Explanation:
Q = m × c × ∆T
Where;
Q = amount of energy (J)
m = mass (grams)
c = specific heat capacity
∆T = change in temperature
m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C
Q = 65 × 4.184 × (150 - 100)
Q = 271.96 × 50
Q = 13598 J
Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.
