Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
Answer:
Actually there is no answer but there is so much theories but people mostly says Bug Bang
Answer:
i would help i just dont know sorry
Explanation:
Answer:
65.21 percent
Explanation:
We are to find the percentage yield
We have this equation,
2no+o2 -->2no2
Such that 1500 kilograms/30grams
= Mno2/46
=1500/30 = mno2/46
We cross multiply from here
1500x46 = 30xMnO2
69000 = 30Mno2
69000/30 = MnO2
2300 = Mno2
The percentage yield would be
1500/2300 *100
= 0.6521 x 100
= 65.21%
This answers the question
