Boron trifluoride gas is collected at 2.0 degree c in an evacuated flask with a measured volume of 15.0 l. when all the gas has been collected, the pressure in the flask is measured to be 0.130 atm. calculate the mass and number of moles of boron trifluoride gas that were collected. be sure your answer has the correct number of significant digits.

Moles of boron trifluoride gas that were collected = 11.6 mol

Mass of boron trifluoride gas that were collected = 787 g

Explanation:

Given that:

Temperature = 2.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (2.0 + 273.15) K = 275.15 K

V = 15.0 L

Pressure = 0.130 atm

Using ideal gas equation as:

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

0.130 atm × 15.0 L = n ×0.0821 L atm/ K mol  × 275.15 K

⇒n = 11.6 mol

Thus, Moles of boron trifluoride gas that were collected = 11.6 mol

Molar mass of boron trifluoride gas = 67.82 g/mol

The formula for the calculation of moles is shown below:

Thus,

Mass of boron trifluoride gas that were collected = 787 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.