Astudent calculated the number of moles of carbon dioxide present in a 100.0 g
sample of a gaseous mixture. if the student determined that there are 0.30 moles of
carbon dioxide present in the sample, what is the percent composition of carbon
dioxide in the sample?

13.2%

Explanation:

Mass of the sample = 100g

MM of CO2 =12+ (16 x 2) = 44g/mol

Number of mole of CO2 present in the sample = 0.3

Mass conc. Of CO2 present = 0.3 x 44 = 13.2g

Percentage of CO2 in the sample = (13.2 / 100) x 100 = 13.2%

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.