You can split the process in two parts:
1) heating the liquid water from 10.1 °C to 25.0 °C , and
2) vaporization of liquid water at constant temperature of 25.0 °C.
For the first part, you use the formula ΔH = m*Cs*ΔT
ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J
For the second part, you use the formula ΔH = n*ΔHvap
Where n is the number of moles, which is calculated using the mass and the molar mass of the water:
n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol
=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J
3) The enthalpy change of the process is the sum of both changes:
ΔH total = 1,874 J + 73,480 J = 75,354 J
Answer: 75,354 J
Answer:
69
Explanation:
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Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
<span>) 8% solution means 8g/100 ml. So 500 ml= 40g. </span>
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