Answer:
Here you can use the Clausis Clayperon equation: ln P1/P2=-Ea/R-(1/T1 - 1/T2)
where P1 is the pressure at standard condition: 760 mm Hg
P2 is the variable we need to solve
Ea is the activation energy, which in this case is delta H vaporisation: 56.9 kJ/mol
R is the gas constant 8.314 J/mol or 8.314 J/mol /1000 to convert to kJ
T1 is the normal boiling point 356.7 C, but converted to Kelvin: 629.85K
T2 is room temperature 25 C, but converted to Kelvin: 298.15 K
Once you plug everything in, you should get 4.29*10^-3 mmHg
Explanation:
All atoms of the same element have the same atomic number
Answer:
A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
(4x^2)x
Kc= -----------
(A-2x)^2
PV=nRT
n/v = P/RT = .745/(0.0821)(334+273) = .01495
To Find the initial molarity of NO2
(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L
Thus:
46(A-2x) + 2x(30) + 32x = .515 g/L
46A-92x+60x+32x = .515
46A=.515
A=.01120 M
Using the total molarity found
(A-2x)+2x+x = .01495 M
A+x=.01495
Plug in A found into the above equation:
.01120+x = .01495
x=.00375
Now Plug A and x into the original Equilibrium Constant Expression:
(4x^2)x
Kc= -----------
(A-2x)^2
Kc = 0.000014
Explanation:
E. Carbon dioxide isn't a product, it's taken in by the plant, oxygen is a product.
Answer:
It is an area that is covered in grasses and wildflowers that receives enough rainfall to support the grassland but not the growth of trees.
Explanation:
