Answer:
(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.
Explanation:
Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:
Q = (pBrCl)²/(pBr₂*pCl₂)
Q = 2²/(1x1)
Q = 4
As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.
The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.
Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.
Not 100% sure, but I'd have to say this:
When rounding for sig figs, you look at the other numbers in your calculation and see which one has the least amount of sig figs. This number is 2.70. (3 sig figs). So, you round to 3 significant figures in your calculation.
Hope this helped!
Answer:
Explanation:
We can use the Ideal Gas Law to calculate the density of the gas.
pV = nRT
n = m/M Substitute for n
pV = (m/M)RT Multiply both sides by M
pVM = mRT Divide both sides by V
pM = (m/V) RT
ρ = m/V Substitute for m/V
pM = ρRT Divide each side by RT
Data:
p = 1.00 bar
M = 49 g/mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 0 °C = 273.15 K
Calculation:
ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L
The density of the gas is .
I think it's letter:
B.Coefficients are placed in front of the reactants and/or products
