A student collects 300mL of hydrogen gas at 22°C and
91.9kPa. What is the temperature of this gas when the
pressure is latm and the volume is 2L? *

2169.1K

Explanation:

Given that,

Original volume of hydrogen gas (V1) = 300mL

[Convert 300mL to liters

If 1000mL = 1L

300mL = 300/1000 = 0.3L]

Original temperature of hydrogen gas (T1) = 22°C

[Convert 22°C to Kelvin by adding 273

22°C + 273 = 295K]

Original pressure of hydrogen gas = 91.9 kPa

[Since new pressure is in atm, convert 91.9 kPa to atmosphere

If 101.325 kPa = 1 atm

91.9 kPa = 91.9/101.325 = 0.9069 atm]

New volume of hydrogen gas (V2) = 2L

New temperature of hydrogen gas (T2) = ?

New pressure of hydrogen gas (P2) = 1 atm

Since pressure, volume and temperature are given, apply the formula for the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(0.9069 atm x 0.3L)/295K = (1 atm x 2L)/T2

0.272 atm•L / 295K = 2 atm•L / T2

To get the value of T2, cross multiply

0.272 atm•L x T2 = 2 atm•L x 295K

0.272 atm•L•T2 = 590 atm•L•K

Divide both sides by 0.272 atm•L

0.272 atm•L•T2/0.272 atm•L = 590 atm•L•K/0.272 atm•L

T2 = 590 atm•L•K / 0.272 atm•L

T2 = 2169.1K

Thus, the new temperature of the hydrogen gas is 2169.1K

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.