Acontainer of what volume would be required to store 60.0 moles of a gas at a pressure of 20.0 atm and a temperature of 50.0 celsius? 12.3 l 79.6 l 8060 l 1250 l

The answer to your question is:  79.6 L

Explanation:

Data:

n = 60 moles

Pressure = P = 20.0 atm

Temperature = T = 50°C

Volume = V = ?

Convert temperature to Kelvin

      °K = 273 + 50

      °K = 323

Formula  (ideal gas law)

               PV = nRT        solve for V

              V = nRT / P

             V = (60)(0.08206)(323) / 20        Substitution

            V = 1590.32 /20                    

             V = 79.6 L

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

the answer for sodium is 0.186

the answer for rubidium is 0.251