Chem Help!!?

Question

I have a test coming up and need help figuring out some review questions!!

*CO gas diffuses through a barrier @ a rate of .280mL/min. If an unknown gas diffuses through the same barrier @ a rate of .277mL/min, what is the molar mass of the gas?

*Excited H atoms emit light in the ultraviolet at 2.47x10^15 Hz. what is the energy of a single photon w/ this freq?

*The energy required to break 1 mole of Cl-Cl bonds in Cl2 is 242kJ/mol. What is the longest wavelength of light capable of breaking a single Cl-Cl bond?

*What is the de Broglie wavelength of a 120g golf ball traveling @ 65km/hr?


THANK YOU SOOOO MUCH!

Answer 1

1) CO gas diffuses through a barrier @ a rate of .280ml/ min, i.e .00466ml/sec. If an unknown gas diffuses through the same barrier @ a rate of .277ml/sec i.e. .00461ml/sec. According to Graham's law of diffusion, the rate at which the gas diffuses is invesely proportional to the vapour density of the gas, i.e r1/r2 =squre root of (vd2/vd1)
As vd=molecular mass /2
r1/r2 = squre root of (M2/M1)
M1= molecular mass of CO = 28
0.00466/0.00461 =squre root of ( M2/28)
1.01^2 =M2/28
M2 = 28.5628


2) Excited H atoms emit light in the ultra-violet region at
2.47x10^15 Hz
frequency = f = 2.47x10^15 Hz
Energy = plank's constant x frequency
= 6.62 x 10^(-34) x 2.47x10^15
= 16.3514 x 10^(-19) J


4) Mass = m = 120 g = 0.12 kg
velocity = v =65km/hr = 1.8m/s
deBrogali's wavelength = plank's constant/mv
= =30.6 x 10^(-34) m

I',m sorry i can't answer the 3rd one.

Answer 2

I can do the Physics type questions-:

2) Frequency = 2.47 x 10^15 Hz

Energy = Planck's Const. x Frequency
Energy = (6.626 x 10^-34) x (2.47 x 10^15)
Energy = 1.6366 x 10^-18 Joules or 10.23 electron Volts

4) De Broglie-:

Convert to SI units.....
120g = 0.120 kg
65 km/hr = 18.056 ms^-1

Lambda = Planck Const. / mass x velocity
Lambda = 6.626 x 10^-34 / 0.120 x 18.056
Lambda = 3.058 x 10^-34 metres
(i.e. This wavelength is too small to measure accurately)

Answer 3

CO has a molecular mass of 28
The formula needed is DCO* (mCO)0.5 = DX *(mX)^0.5
D =diffusivity , mX=molecular weight X

so (mX)^0.5= DCO* (mCO)0.5/DX =(28)^0.5*0.280^0.5/0.277
mX=28.6

the energy of the photon is
E=hf E=6.6210^-34*2.4710^15=1.630^-18J = 10.2 eV

Bond Cl-Cl E =242kJ/mol
for 1 molecule E=242000/6.0210^23=4.0210^-19J
E = hc/L so L = hc/E= 6.6210^-34*310^8/4.0210^-19=4.9410^-7m = 494nm (visible)

Golf ball wavelength = h/mv= 6.6210^-34/(0.12*v)

v in m/s = 65000/3600=18m/s

wavelength = 6.6210^-34 /(0.12*18) =3*10^-34m