How many grams of pentane are contained in a vapor sample that occupies a volume of 0.285 l at 57°c and 1.0 atm? .

0.260 g
0.515 g
0.760 g
1.00 g

The answer to your question is: mass = 0.760 g

Explanation:

Data

V = 0.285 L

T = 57°C = 330°K

P = 1 atm

R = 0.082 atm L/mol°K

mass = ?

Formula

                PV = nRT

                n = PV / RT

Substitution

                n = (1)(0.285) / (0.082)(330)

                n = 0.285 / 27.06

                n = 0.011

mW Pentane = (12 x 5) + (12 x 1) = 72 g

                                 72 g of pentane 1 mol

                                   x                         0.0105 mol

                                  x = (0,0105 x 72) / 1

                                  x = 0.756 g ≈ 0.760 g

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.