Answer:
See below
Explanation:
<u> Name </u> <u>Formula </u> <u> Major species </u> <u> </u>
Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),
Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)
Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)
Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)
Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)
- Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
- The compounds containing metals are ionic. They produce ions in solution.
- ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
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Answer:
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B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.
Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P).
