Verified answer

Balanced equation:

2Nb +H2 → 2NbH

2mol Nb react with 1mol H2 to produce 2 mol NbH

Molar mass Nb = 92.9064 g/mol

0.78g Nb = 0.78/92.9064 = 8.4*10^-3 mol Nb

This will react with 4.2*10^-3 mol H2

Mol of H2 started with:

PV = nRT

6.33*0.028 = n*0.082057*298

n = 7.25*10^-3 mol

Mol of H2 unreacted = 7.25*10^-3 - 4.2*10^-3 = 3.048*10^-3 mol H2

You finally have 3.048*10^-3 mol H2 in 28ml container at 25°C

Calcualte pressure by gas law equation:

PV = nRT

P * 0.028 = (3.048*10^-3) * 0.082057 * 298

P = 2.66 atm

At equilibrium the partial pressures of the hydrogen and iodine gases blended would be equivalent to 18 atm on condition that it particularly is the partial presure of the hi at equilibrium. initially the stress of the H2 is 15/25 or 3/5 of the entire partial stress, so the H2 would be 3/5 or 0.6 x 18 = 10.8 atm and the partial stress of the I2 at equilibrium would be 0.4 x 18 = 7.2 atm

This question seems a litle tricky since they are not giving you the final temperature or the final volume. So we're gonna asume that the temperature and volume remain constant.

For this question you re gonna use the formula:

P1V1/T1 = P2V2/T2

P = pressure in atm

V = Volume in Liters

T = Temperature in Kelvin,

So they are giving the volume in mL so you must change it to liters.

28.0mL * 1L / 1000mL = 0.028 L

And the temperature is i celsius so add 273 to get it in Kelvin.

25 + 273 = 298 K

Then you move around the formula so you are solving for P2, All the 2 values are the final values.

So you would have:

P2 = P1V1T2 / T1V2 And now we just plug in.

(6.33 atm ) (0.028L)(298K) / (298K)(0.028L) = 6.33 atm

So the pressure would be the same,

Hope that helps!!!!

67 psi