The equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant.
Depending on the concentration of titrant we could be adding little excess of it and this may result in persistence of color of solution. After continuous stirring for a while the excess titrant may react with dissolved CO₂ in air and thus decolorizing the solution.
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Answer:
Kc = 9.52.
Explanation:
<em>A + 2B ⇌ C,</em>
Kc = [C]/[A][B]²,
Concentration: [A] [B] [C]
At start: 0.3 M 1.05 M 0.55 M
At equilibrium: 0.3 - x 1.05 - 2x 0.55 + x
0.14 M 1.05 - 2x 0.71 M
- For the concentration of [A]:
∵ 0.3 M - x = 0.14 M.
∴ x = 0.3 M - 0.14 M = 0.16 M.
∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.
<em>∵ Kc = [C]/[A][B]²</em>
∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.
Answer: The balanced equation for overall reaction is:
(CH3)3AuPH3 ----> C2H6 + (CH3)AuPH3
Explanation:
The reaction mechanism is given as follows:
Step 1: (CH3)3AuPH3 ↔ (CH3)3Au + PH3 (fast)
Step 2: (CH3)3Au → C2H6 + (CH3)Au (slow)
Step 3: (CH3)Au + PH3 → (CH3)AuPH3 (fast)
To balance this equation, firstly, we conduct proper atom count for each steps of the reaction mechanism.
It is important to note that for a reaction that involves several steps, the rate law is normally derived from the slow step ( which is step2 from the above mechanism).
Therefore, the balanced chemical equation for the overall reaction is:
(CH3)3AuPH3 ----> C2H6 + (CH3)AuPH3
<span>I think it is a but dont take me for it. I searched it and thats what I found since liquid in beaker B is more dense then realtivley speaking the object is less dense compared other liquid and floats at a higher level.</span>
Red blood cell is the answer I think
