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3 years ago

What are the four functions of cell division?

Chemistry

2 answers:

Softa [21]3 years ago

6 0

Answer:

Cellular division has three main functions: (1) the reproduction of an entire unicellular organism, (2) the growth and repair of tissues in multicellular animals, and (3) the formation of gametes (eggs and sperm) for sexual reproduction in multicellular animals.

Marat540 [252]3 years ago

3 0

Answer:

There are four key reasons why a cell may be required to divide mitotically:

Tissue repair / replacement.

Organismal growth.

Asexual reproduction.

Development (of embryos)

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If a gas is measured to have a pressure of 1.2 atm at 227 K, and its origina

ICE Princess25 [194]

Answer:

1.05

Explanation:

P1 over t1 is equal to P2 over T2, then solve for X

X/199= 1.2/227

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2 years ago

A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin

Marysya12 [62]

First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V

In order to determine the potential of the cell, we find the difference between the two values. For this:

E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V

The second option is correct. (The difference in values is due to different values in literature, and it is negligible)

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2 years ago

At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti

mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0

3 years ago

If anyone can get this question, 20 points for you!

torisob [31]

Answer:

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Explanation:

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2 years ago

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Vladimir79 [104]

Answer:

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Explanation:

I think thats the answer in your question

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2 years ago