Answer:
1. B.10 mol/l
2. D.
3.A.
Explanation:
Totoo po yan ganyan po yong sakin
Answer:
Ka = 
Explanation:
Initial concentration of weak acid =
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:

(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)

0.000000135 <<< 0.00045

Answer:

Explanation:
We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.
We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.
Next, convert these masses to moles, using the molar masses found on the Periodic Table.
- P: 30.974 g/mol
- Cl: 35.45 g/mol
Use the molar masses as ratios and multiply by the number of grams. 

Divide both of the moles by the smallest number of moles to find the mole ratio.


The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as:<u> PCl₃</u>
Answer:
Kindly check the explanation section.
Explanation:
From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.
In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;
∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).
That is the heat of formation of products minus the heat of formation of the reaction g specie(s).
Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.
−382.185 = x - 472.435kj/mol = 90.25 kJ/mol