The temperature of a sample of gas in a steel tank at 0.30 ATM at 25 Celsius if the pressure W20.59 ATM what would be the final temperature inside the tank in kelvin

The final temperature : 313.06 °C

Gay Lussac's Law  

When the volume is not changed, the gas pressure is proportional to its absolute temperature  

P1=0.3 atm

T1=25 + 273=298 K

P2=0.59 atm

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.