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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!!!

You might be interested in
Which equation is correctly balanced?
ludmilkaskok [199]

Answer:

C) 2 H₂ + O₂  →  2 H₂O

Explanation:

4 atoms of hydrogen on reactant side

2 atoms of oxygen on reactant side

4 atoms of hydrogen on product side

2 atoms of oxygen on product side

6 0
3 years ago
The volume of oxygen, collected over water, is 185 mL at 25 degrees Celsius and 600 torr. calculate the dry volume of the oxygen
ivolga24 [154]

Answer:

0.1593 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.

P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.

<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>

4 0
3 years ago
Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane
dolphi86 [110]

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center  of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has  chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as :  methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group  is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

#SPJ4

8 0
1 year ago
Sulfuric acid reacts with aluminum hydroxide by double replacement. if 34 g of sulfuric acid react with 33 g of aluminum hydroxi
Lina20 [59]
Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.

4 0
3 years ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0
3 years ago
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