The reaction below virtually goes to completion because cyanide ion forms very stable complexes with Ni2+ ion:[Ni(H2O)6]2+(aq) +
4 CN-(aq) → [Ni(CN)4]2-(aq) + 6 H2O(l)At the same time, incorporation of 14C labelled cyanide ion (14CN-) is very rapid:[Ni(CN)4]2-(aq) + 4 14CN-(aq) = [Ni(14CN)4]2-(aq) + 4 CN-(aq)Which statement below is correct with regard to stability and rate of reaction?
1 answer:
Answer:
Option A and D are correct.
Unstable species react rapidly.
Stable species do not react rapidly.
Explanation:
The complete question is attached to this solution.
The more stable a reactant is, the less reactive it will be. A stable reactant has a very stable structure in which it will avoid any perturbations. And for a reaction to occur, the bonds in the reactant must break down to form the products. A stable reactant has very strong bonds that aren't easy to break down, hence, reactions involving very stable reactants do not proceed rapidly.
And the more unstable a reactant specie is, the more rapidly it reacts. This is why the reaction involving the less stable isotope of carbon; Carbon-14 is very rapid. It is the same reason as explained above that is responsible for this. The bond between unstable species are not strong and are easily breakable, thereby leading to a quick reaction.
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Answer:
431.38 mg protein / mL
Explanation:
This is an example of the <em>Kjeldahl method</em>, for nitrogen determination. All nitrogen atoms in the protein were converted to NH₃ which then reacted with a <u>known excess of HCl</u>. This excess was later quantified via titration with NaOH.
First we calculate the <u>total amount of H⁺ moles from HCl</u>:
- 0.0388 M HCl * 10.00 mL = 0.388 mmol H⁺
Now we calculate the <u>excess moles of H⁺</u> (the moles that didn't react with NH₃ from the protein), from the <u>titration with NaOH</u>:
- HCl + NaOH → H₂O + Na⁺ + Cl⁻
- 0.0196 M * 3.83 mL = 0.075068 mmol OH⁻ = 0.0751 mmol H⁺
Now we substract the moles of H⁺ that reacted with NaOH, from the total number of moles, and the result is the <u>moles of H⁺ that reacted with NH₃ from the protein</u>:
- HCl + NH₃ → NH₄⁺ + Cl⁻
- 0.388 mmol H⁺ - 0.0751 mmol H⁺ = 0.313 mmol H⁺ = 0.313 mmol NH₃
With the moles of NH₃ we know the moles of N, then we can <u>calculate the mass of N</u> present in the aliquot:
- 0.313 mmol NH₃ = 0.313 mmol N
- 0.313 mmol N * 14 mg/mmol = 4.382 mg N
From the exercise we're given the concentration of N in the protein, so now we <u>calculate the mass of protein</u>:
- 4.382 mg * 100/15.7 = 27.91 mg protein
Finally we <u>calculate the protein concentration in mg/m</u>L, <em>assuming your question is in 647 μL</em>, we first convert that value into mL:
- 647 μL *
0.647 mL
- 27.91 mg / 0.647 mL = 431.38 mg/mL