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Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:
Number of moles (n)
can be written as:
where, m = given mass
M = molar mass
where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant =
T = temperature of the gas =
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
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