How does brock answer the charge that euthanasianis the deliberating killing of na innocent person?
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Answer:
4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.
Explanation:
The cytochromes are <u>proteins that contain heme prosthetic groups</u>. Cytochromes <u>undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome</u>:
The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes <em>b</em> and <em>c₁</em>, and a Fe-S center. This second complex reduces cytochrome <em>c</em>, a water-soluble membrane peripheral protein. Cytochrome <em>c</em>, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes <em>a</em>, <em>a₃</em> and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.
Each transporter "downstream" is <u>more electronegative</u><u> than its neighbor </u>"upstream"; oxygen is located in the inferior part of the chain. Thus, the <u>electrons fall in an energetic gradient</u> in the electron chain transport to a more stable localization in the <u>electronegative oxygen atom</u>.
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
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In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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