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:
im guessing it's the second one
A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH. In variable form, a thermochemical equation would look like this:
A + B → CΔH = (±) #
Where {A, B, C} are the usual agents of a chemical equation with coefficients and “(±) #” is a positive or negative numerical value, usually with units of kJ.
please mark as brainliest
Answer:
4960000000 pm
Explanation:
4.96*1000000000= 4960000000
