Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
Mixed economy. Wherever the private and public sector have an active role, they are determined to be a mixed economy.
Answer:
the third option
Explanation:
The circuit is incomplete and broken, and the lights in the circuit do not shine.
An open circuit means an incomplete circuit.
The first option is wrong since the lights do not shine in an open circuit.
The second option is wrong since an open circuit is incomplete.
The fourth option is also wrong since an open circuit is broken or incomplete.
∴ The third option is correct.
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
Answer:
23.55 L
Explanation:
USe the following 'identity' of gs laws
P1 V 1 / T1 = P2 V2 / T2 ( T must be in Kelvin)
re arrange to
P1 V 1 T2 / (T1 P2) = V2 NOW SUB IN THE VALUES
752 * 34.2 * ( 34 + 273.15) / [( 229 + 273.15) * 668] = V2 = 23.55 L
