(p1)(V1)/(T1) = (p2)(V2)/(T2)
(1.00 atm)(V) / (273 + 25K) = (40.0 atm)(V/10) / (273 + T)
273 + T = (40.0)(1/10)(273 + 25K) / (1.00)
T = 919°C
Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Explanation:
1) Data:
Water ⇒ C = 1 cal/g°C
m = 65.8 g
Ti = 31.5°C
Tf = 36.9°C
Heat, Q = ?
2) Formula:
Q = mCΔT
3) Calculations:
Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal
4) You can convert from calories to Joules using the conversion factor:
1 cal = 4.18 J
⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J
Answer:

Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

Best regards.
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