The normal boiling point<span> of </span>ethanol<span> is 78.4 degrees C and, at thistemperature, </span>the vapor pressure<span> is 101325 Pascals (Pa) or 760manometric units
thx hope this helped bye.</span>
1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa
101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm
1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg
B. Holocaust
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Answer:
119.5 J
Explanation:
First we <u>calculate the temperature difference</u>:
- ΔT = 100 °C - 50 °C = 50 °C
Then we can <u>calculate the heat released</u> by using the following formula:
Where q is the heat, Cp is the specific heat, ΔT is the temperature difference and m is the mass.
We <u>input the data</u>:
- q = 0.239 J/g°C * 50 °C * 10.0 g
