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
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
I think it’s , D. This can cause sonic booms that can damage building structures
Answer:
The mass percent of potassium is 39%
Option C is correct
Explanation:
Step 1: Data given
Atomic mass of K = 39.10 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate molar mass of KHCO3
Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0
Molar mass KHCO3 = 100.12 g/mol
Step 3: Calculate mass percent of potassium (K)
%K = (atomic mass of K / molar mass of KHCO3) * 100%
%K = (39.10 / 100.12) * 100%
%K = 39.05 %
The mass percent of potassium is 39%
Option C is correct
