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:
I = 1.23 A
Explanation:
Given that,
The resistance of the lightbulb, R = 96.8 Ω
Voltage, V = 120 V
We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.

So, the current flows through the bulb is 1.23 A.
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Answer:
24.309
Explanation:
mass×percentage+.. = average amu
23.985×78.7/100+24.986×10.13/100+25.983×11.17/100
to three decimal places
=24.309