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sattari [20]
3 years ago
15

Which best describes the Tyndall effect?

Chemistry
2 answers:
konstantin123 [22]3 years ago
7 0
<span>The correct answer is the scattering of light by particles in a mixture. Even more correct would be in a coloid or even sometimes in a suspension. What this means is that it would be when light particles scatter around things like mud because mud is a suspension, or if there was flour in water then that.</span>
8_murik_8 [283]3 years ago
6 0

Answer:

The scattering of light by particles in a mixture.

Explanation:

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After finding the oxidation states of atoms, you identify the half reactions (option c).

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a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

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