Answer:
2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)
Explanation:
A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.
The only equation that meets those conditions is the last one.
A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.
C is wrong because Li⁺ and SO₄²⁻ are not elements.
D is wrong because it shows the formation of 8 mol of Li₂SO₄.
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span>
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration
Answer:
first we add the same direction. 12N + 32 N=44N .
then we add the forces. 54 up + 44N down= 10N up
