Answer:25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
Explanation:
Answer:
60 J
Explanation:
The law of conservation of energy states that energy is neither created nor destroyed, just converted into different forms. This means the total mechanical energy of the object at point A will be the same as the total mechanical energy at point B, and the question tells us the total of that mechanical energy is 150 J. Note we are assuming no energy is lost from the system as heat.
At point B, if the potential energy is 90 J, the remainder of the 150 J total must be kinetic energy. KE = 150 J - 90 J = 60 J.
I believe the correct answer is C, but I'm 100% on this. Hope this helped though!
-TTL
The Kj of heat that are needed to completely vaporize 1.30 moles of H2O if the heat of vaporization for water is 40.6 Kj/mole is calculated as below
Q(heat) = moles x heat of vaporization)
=1.30 mol x40.6 kj/mol= 52.78 Kj is needed
