Because more than one substance was released (following a color change signifying a chemical reaction), the sample was indeed, a compound.
Answer:
[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22
Explanation:
Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵
M (molar mass) of BA (Benzoic Acid) = 122 g/mol
Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M
We should consider the equation once it reaches the equilibrium:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
C - x x x
And, for the Kₐ:
Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³
Then: x² + Kₐx - KₐC = 0
x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0
Resolving this cuadratic equation (remember to use Baskara equation), we obtain:
x = 6.083x10⁻⁴ M
Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M
[C₆H₅COOH] = C - x = 3.98x10⁻³ M
pH = -Log [H⁺] = 3.22
Hello!
This is a difficultly easy question, mostly because it seems like all of the answers would be right, but if I had to choose one, I would say (D) because this option gives the most detail.
Option A only tells us how fast a car was moving
Option B only tells us the rate the water falls and how fast it is going
Option C only tells how far and not the speed or destination
Option D is correct, it tells us the speed/rate, the destination, and what is moving
Therefore, Option (D) is correct
If this helps, please mark brainliest, have a great day!!
CH₃-CH=CH₂ propene
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