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sergey [27]
3 years ago
9

Explain what is cellular growth and repair? Why is it important?

Chemistry
1 answer:
HACTEHA [7]3 years ago
3 0

Answer:

Cell growth usually refers to cell proliferation, the increase in cell numbers that occurs through repeated cell division. Cell growth can also refer to the enlargement of cell volume, which can take place in the absence of cell division. As living things grow, some cells die or become damaged and need replacements. Some single-celled organisms use a type of mitosis as their only form of reproduction. In multicellular organisms, cell division allows individuals to grow and change by expanding the number of total cells.

Hope this helps!!!

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12. What is the mass, if the density is 2 and the volume is 6? *<br> 3<br> O 12<br> O 15
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Convert the following: 100 hg(hectograms) to g(grams)
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Answer:

A

Explanation:

10,000g

just multiply the amount of hectograms by 100 to get grams

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5 0
3 years ago
Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
n=3

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
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