Answer:
There are 29.4 grams of oxygen in the container
Explanation:
<u>Step 1: </u>Data given
Volume = 20.0 L
Pressure = 845 mmHg
Temperature = 22.0 °C
Molar mass of O2 = 32 g/mol
<u>Step 2:</u> Ideal gas law
p*V = n*R*T
⇒ p = the pressure of the gas = 845 mmHg = 1.11184
⇒ V = the volume of the gas = 20.0 L
⇒ n = the number of moles = TO BE DETERMINED
⇒R = the gasconstant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 22°C + 273 = 295 Kelvin
n = (p*V)/(R*T)
n = (1.11184*20.0)/(0.08206*295)
n = 0.9186 moles
<u>Step 3:</u> Calculate mass of NO2
Mass of O2 = Moles O2 * Molar mass O2
Mass of O2 = 0.9186 moles * 32 g/mol
Mass of O2 = 29.4 grams
There are 29.4 grams of oxygen in the container
25 x 10.5 = 262.5
So the approx. is 260 g which is answer D.
This is so that all measurements can be used without having to calculate from different units.
Answer:
temporary changes are the changes which are there only for a short period if time.
Explanation:
generally temporary changes are reversible. permanent changes are the changes which remain for a longer time and are not reversible
Answer:
=3 means is 3 or greater so that would be f and g subshells
=0 means is 0 or greater so that would be s, p, d, f and g subshells
=1 means is 1 or greater so that would be p, d, f, and g subshells
=4 means is 4 or greater so that would be g only
