Answer : 5100 KJ
Explanation : In the given reaction of

we can see that 2 moles of Al is producing -850 KJ of energy
so for 12 moles of Al we can calculate as;
12 moles of Al X

= -
5100 KJ
So, we can say that 12 moles of Al will need -5100 KJ of energy.
<h3><u>Answer;</u></h3>
pOH = 3.08
<h3><u>Explanation;</u></h3>
NX3 + H2O <----> NHX3+ + OH-
Kb = 4.0 x 10^-6
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃)
x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = c(OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
<u>= 3.08</u>
Answer:

Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.4%7D%3D3.98x10%5E%7B-4%7D)
And the percent ionization is:
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20%5C%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
We compute the concentration of the acid, HA:
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5C%25%20%5C%20ionization%7D%2A100%5C%25%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%7D%7B66%5C%25%7D%20%20%2A100%5C%25%5C%5C%5C%5C)
![[HA]=6.03x10^{-4}](https://tex.z-dn.net/?f=%5BHA%5D%3D6.03x10%5E%7B-4%7D)
Thus, the Ka is:
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%2A3.98x10%5E%7B-4%7D%7D%7B6.03x10%5E%7B-4%7D%7D%5C%5C%20%20%5C%5CKa%3D2.63x10%5E%7B-4%7D)
So the pKa is:

Regards.
A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall. How tall is the actual car?
<h2>Answers:</h2>
<h3>A. 120 cm</h3>
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