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Vlada [557]
3 years ago
13

Why doesn't cell division result in two cells with half the needed number of chromosomes?

Chemistry
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

This is because the number of chromosomes in a cell always remain fixed or constant.

Explanation:

During cell cycle the DNA of the cell is replicated in the S phase resulting in the formation of 2 daughter DNA molecules which are distributed in the two newly formed daughter cells.Thus chromosome is replicated once per cycle and chromosome number remains fixed in each and every cell.

      According to the question if cell undergo division in two cells  with half the needed number of chromosome then after few cell divisions no chromosome will be left for distribution and the cell will die.

 

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The energy transformation occurs in a gasoline-powered car is c<span>hemical energy is converted into mechanical energy. The answer is letter A. The rest of the choices do not answer the question above.</span>
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. In which reaction is nitric acid acting as an oxidising agent?
Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

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What does the position of an element in the periodic table tell you?
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3 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

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Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

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3 years ago
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