Gas laws, ideal gas law, Grahms Law

[Hitman47]

Im having trouble setting up these word problems, i cant visualize it. Can you guys teach me how to do these

 

1. Calculate the molecular weight of a gas if 4.5L of the gas at 785 torrs and 23.5 Celcius weigh 13.5g.

 

2.) 0.453 mol of a gas confined to a 15.0L container exerts a pressure of 1.24atm on the walls of the container. What is the temperature of the gas?

 

3) 2.0 * 10^-5 g of hydrogen gas at 155 Celcius exert a pressure of 322.5 torr on the walls of a small cylindrical tube. What is the volume of the tube?

 

4) If 250mL of nitrogen are collected over water at 25 Celcius and 750 torr, what will the volume of the dry gas be at STP?

 

5) A certain gas is collected over water at 740mm and 23 Celcius. The collecting tube is left in place, and the volue is not measured until the next day when the pressure is 745 mm and the temperature 20.0 Celcius, at which time the volume is found to be 15.3mL. What was the original volume?

 

 

 

 

 

 

And grahams law i dont get

 

 

When two cotton plugs, one moistened with ammonia and the other with the hydrobromic acid, are simultaneously inserted into opposite ends of a glass tube 87.0 cm long, a white ring of NH4Br forms where Gaseous NH3 and gaseous HBr first come into contact

 

NH3 (g) + Hbr ----> NH4Br(s)

 

At what distance from the ammonia-moistened plug does this occur

 

 

 

 

 

 

 

 

 

 

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sorry if it is too much, but i dont know how to do this

 

 

 

Thanks for any help

[wolfson]

Gas Laws:

 

Ok here I will try to answer your questions:

 

(1). n = PV / RT n = [ (785.0 mmHg / 760.0 mmHg atm-1) (4.50L) ] / [ (0.08206 L atm mol-1 k-1) (296.5K) ]

 

Then, divide the grams given (13.5g) by the moles just calculated above. This will be the molecular weight.

 

(2). T = (PV) / (nR) P =125.643Kpc, V=15.0L, n=0.453, R = 8.314 J K mol-1

 

T= (125.643x15) / (0.453x8.314) = 500.40K.

 

(3) V = (nRT) / P n=0.001mol-1, R=8.314 J K mol-1, T=428 K, P=42.93Kpc.

 

V = (0.001x8.314x428) / 42.93 = 0.0829 x 1000 =82.89ml.

 

(4) V = (nRT) / P n=0.250mol-1 R=8.314 J K mol-1, T=273K, P=101.325

 

V= (0.250x8.314x273) / 101.325 = 560ml.

 

 

(5) V = (nRT) / P (n?x8.314x293) / 99.33Kpc = 15.30ml a decreasment of 5mmHg and 3 degrees so at each degree there is 0.765ml so 23 degree¡¯s would be 0.765 x 23 = 17.60ml, 17.60ml was the original volume.

 

 

(6) Length of tube = 87.0cm (870mm)

where m=mass of molecule and c = speed of molecule so that:

KE (HBr(g)) = KE (NH3(g))

(1/2) (mHCl) (cHBr)2=(1/2) (mNH3) (cNH3)2

(mHBr) / (mNH3) = (cNH3)^2 / (cHBr)^2)

[ (cNH3) / (cHBr) ] = ¡Ì [ (mHBr) / (mNH3) ]

Since speed = distance/time, and the time is the same for both kinds of molecules in this experiment,

[ Distance NH3 / Distance HBr ] = ¡Ì [ (mHBr) / (mNH3) ]

 

Distance traveled by NH3 = 51.30cm

Distance traveled by Hbr = 87.0 ¨C 51.30 =35.7cm

Distance met = 51.30cm from NH3.

[Dudde]

I didn't calculate them out myself to check wolfson's answers, but from what I scanned he looks correct