I believe that would be oil and vinegar.
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
The vapor pressure of the jar with a large amount of water and the jar with a small amount of water is the same. Vapor pressure is an intrinsic property. This means that this property is not dependent on the amount of the substance. Vapor pressure is dependent only on temperature and since the temperature is the same for both jars, their vapor pressures are also the same.
With all of the information given (pressure, volume, temperature, and the molar mass), we need a formula that relates this all together. The formula we need is the ideal gas law, PV=nRT. Since the pressure is defined in millimeters of mercury, we need the R value that correlates with this, which is 62.4; on top of this, we need the temperature in Kelvin - simply add 273.15 to convert from Celsius. With all of this information, simply plug-and-chug:
PV=nRT
(800)(3.7) = n(62.4)(37 + 273.15)
n = 0.1529 moles
Finally, the problem is asking the amount of air in grams. We have moles, so all we need to do is multiply that value by the molar mass.
0.1529 moles x 29 grams per mole =
4.435 grams of air
The balloon has 4.435 grams of air inside it.
Hope this helps!