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SCORPION-xisa [38]
3 years ago
5

Which organisms have greater variation and produce offspring through asexual reproduction?

Chemistry
1 answer:
Sedaia [141]3 years ago
8 0

Answer: Fungi

Explanation:

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Explain the following: Unless the loss of reactant is extreme, only the loss of limiting reactant mass truly introduces error in
natita [175]

Answer:

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Explanation:

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How does the ocean current affect the air temperature in japan.
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Draw the mechanism for the addition of methylmagnesium bromide to propanoyl chloride (CH3CH2COCl) to give 2-methyl-2-butanol.
Ghella [55]

Answer:

Reaction mechanism has been given below

Explanation:

  • Propanoyl chloride contains an electrophilic carbonyl center to give  nucleophilic acyl substitution reaction by removing Cl group with a strong nucleophile like methylmagnesium bromide.
  • Two equivalent of methyl magnesium bromide react with cabonyl center followed by protonation of negatively charged oxygen atom gives 2-methyl-2-butanol.
  • In the first step, nucleophilic acyl substituion occurs. In the second step, nucleophilic addition occurs. In the third step, protonation occurs.
  • Reaction mechanism has been shown below.

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MULTIPLE CHOICE QUESTION
german

Answer:

radiation

Explanation:

8 0
3 years ago
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A solution is made by dissolving 0.541 mol of nonelectrolyte solute in 865 g of benzene. Calculate the freezing point, T f , and
Kazeer [188]

Answer:

T f = (Normal freezing point of benzene - Molality of the solution × freezing point depression per molal of benzene) = 5.5 ° C - .625 m × 5.12° C = 2.30° C.

T f = 2.30° C.

T b = (Normal boiling point of benzene - Molality of the solution × boiling point elevation per molal of benzene) = 80.1 ° C. - .625 × 2.53° C = 81.68 ° C.

T b = 81.68 ° C.

Explanation:

The molality of the solution is .541 ÷ .865 = .625 m.

Ideally benzene freezes at 5.5 ° C and the depression of the freezing point is 5.12° C per molal.

At a molality of 0.625 the freezing point will be depressed by (molality of solution × freezing point depression per molal of benzene) or .625 × 5.12° C = 3.20° C.  So the resulting freezing point will be (Ideal freezing point - freezing point depression) or T f =5.5 ° C. - 3.20 ° C. = 2.30° C.

Also ideally benzene boils at a temperature of 80.1° C and the elevation of the boiling point per molal is 2.53° C .  So the boiling point of benzene will be elevated by (molality of solution × boiling point elevation per molal of benzene) .625 × 2.53° C = 1.58° C.  So the boiling point of the current solution, T b, will be 80.1 ° C. + 1.58 ° C. = 81.68 ° C.

8 0
3 years ago
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