Reliable results are results that can be... A: Communicated
Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is 
.
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas = 
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is .
Divide the mass of the compound in grams by the molar mass you just calculated. The answer is the number of moles of that mass of compound. For example, 25 grams of water equals 25/18.016 or 1.39 moles.
Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).
We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.
Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.
(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).
Just look it up on goog^le or a chart
