Answer:
pH = 11.216.
Explanation:
Hello there!
In this case, according to the ionization of ammonia in aqueous solution:
We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D%20%5C%5C%5C%5C1.80x10%5E%7B-5%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.150-x%7D)
However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:
Which is also:
![[OH^-]=1.643x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.643x10%5E%7B-3%7DM)
Thereafter we can compute the pOH first:
Finally, the pH turns out:
Regards!
Answer:
The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.
Explanation:
Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.
Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).
As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.
As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.
The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.
Answer:
Q = 5555.6J
Explanation:
Mass of glass piece, m = 453g
initial temperature = 25.7°C
temperature to be attained = 40.3°C
⇒change in temperature, Δt = 40.3 - 25.7 = 14.6°C
specific heat of glass, s = 0.840J/g°C
Heat absorbed, Q = msΔt
⇒Q = 453×0.840×14.6 = 5555.592J
⇒<u>Q = 5555.6J</u> (rounded to the nearest tenth)
