This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Answer:
14.93 g
Explanation:
First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:
- 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As
Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:
- 0.199 mol * 74.92 g/mol = 14.93 g
Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.
The answer is double displacement is the answer
