.If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant, what final pressure would result if the original pressure was 750.0 mm Hg? Round to the nearest tenth. Don't forget the units.
what should i plug in?
Plug in the numbers. What else?
P1/T1 = P2/T2
If you have a problem identifying the numbers to use in each position, please detail exactly what you don't undertsand about the process.
If you burn 47.4 grams hydrogen and produce 423 grams water, how much oxygen reacted?
To solve this problem, you can use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the volume is kept constant, we can simplify the equation to P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Given:
P1 = 750.0 mm Hg (you may need to convert this to another unit later)
T1 = 323.0 K
T2 = 273.15 K
To find P2, plug in the values into the equation and solve for P2:
(750.0 mm Hg) / (323.0 K) = P2 / (273.15 K)
To ensure consistency, we need to convert the pressure from mm Hg to the same unit as P2. Let's convert it to atmospheres (atm):
1 atm = 760 mm Hg
750.0 mm Hg = 750.0 mm Hg * (1 atm / 760 mm Hg) = 0.9868 atm
Now, substitute the values into the equation and solve for P2:
(0.9868 atm) / (323.0 K) = P2 / (273.15 K)
Cross-multiply and solve for P2:
P2 = (0.9868 atm * 273.15 K) / 323.0 K
P2 = 0.8350 atm (rounded to the nearest tenth)
Therefore, the final pressure would be approximately 0.8 atm.