Answer:
For the 4p orbital, n=4, which is just the energy level of the orbital. All p orbital has an l quantum number equal to 1, so for the 4p orbital, l=1.
Explanation:
Answer:
Kc = 0.5951 (4 sig. figs.)
Explanation:
For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)
ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc
=> lnKc = - ΔG°/R·T
ΔG° = +12.86 Kj/mol
R = 8.314 Kj/mol·K
T = 298K
lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹
Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)
Answer:
There must be an equal amount of each element on both sides of the equation.
Hope this helps
good luck
Answer:
Option A; V = 2.92 L
Explanation:
If we assume a lot of things, like:
The gas is an ideal gas.
The temperature is constant.
The gas does not interchange mass with the environment.
Then we have the relation:
P*V = n*R*T = constant.
Where:
P = pressure
V = volume
n = number of moles
R = constant of the ideal gas
T = temperature.
We know that when P = 0.55 atm, the volume is 5.31 L
Then:
(0.55 atm)*(5.31 L) = constant
Now, when the gas is at standard pressure ( P = 1 atm)
We still have the relation:
P*V = constant = (0.55 atm)*(5.31 L)
(1 atm)*V = (0.55 atm)*(5.31 L)
Now we only need to solve this for V.
V = (0.55 atm/ 1 atm)*(5.31 L) = 2.92 L
V = 2.92 L
Then the correct option is A.
