What is the density of aluminum chloride vapor (AlCl3) at 498 K and 0.939 atm? 1.24 g/L
3.06 g/L
5.17 g/L
7.42 g/L

B  3.06 g/L

Explanation:

formula for density is d = PM / RT

1. find given: P(pressure) = 0.939 atm, M(molar mass) = 133.33, R = 0.0821, T(temperature) = 498 K

2. plug in:  d = 0.939*133.33 / 0.0821*498 = 3.0621113 which rounds to 3.06 g/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.