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astraxan [27]
3 years ago
7

5. Determine Differences What are three differences between plant cells and

Chemistry
2 answers:
Anestetic [448]3 years ago
8 0

Alright so...

Size

Animal cells are generally smaller than plant cells. Animal cells range from 10 to 30 micrometers in length, while plant cells range from 10 and 100 micrometers in length.

Shape

Animal cells come in various sizes and tend to have round or irregular shapes. Plant cells are more similar in size and are typically rectangular or cube shaped.

Energy Storage

Animals cells store energy in the form of the complex carbohydrate glycogen. Plant cells store energy as starch.

Proteins

Of the 20 amino acids needed to produce proteins, only 10 can be produced naturally in animal cells. The other so-called essential amino acids must be acquired through diet. Plants are capable of synthesizing all 20 amino acids.

Differentiation

In animal cells, only stem cells are capable of converting to other cell types. Most plant cell types are capable of differentiation.

Growth

Animal cells increase in size by increasing in cell numbers. Plant cells mainly increase cell size by becoming larger. They grow by absorbing more water into the central vacuole.

Cell Wall

Animal cells do not have a cell wall but have a cell membrane. Plant cells have a cell wall composed of cellulose as well as a cell membrane.

Centrioles

Animal cells contain these cylindrical structures that organize the assembly of microtubules during cell division. Plant cells do not typically contain centrioles.

Cilia

Cilia are found in animal cells but not usually in plant cells. Cilia are microtubules that aid in cellular locomotion.

Cytokinesis

Cytokinesis, the division of the cytoplasm during cell division, occurs in animal cells when a cleavage furrow forms that pinches the cell membrane in half. In plant cell cytokinesis, a cell plate is constructed that divides the cell.

Glyoxysomes

These structures are not found in animal cells but are present in plant cells. Glyoxysomes help to degrade lipids, particularly in germinating seeds, for the production of sugar.

Lysosomes

Animal cells possess lysosomes which contain enzymes that digest cellular macromolecules. Plant cells rarely contain lysosomes as the plant vacuole handles molecule degradation.

Plastids

Animal cells do not have plastids. Plant cells contain plastids such as chloroplasts, which are needed for photosynthesis.

Plasmodesmata

Animal cells do not have plasmodesmata. Plant cells have plasmodesmata, which are pores between plant cell walls that allow molecules and communication signals to pass between individual plant cells.

Vacuole

Animal cells may have many small vacuoles. Plant cells have a large central vacuole that can occupy up to 90% of the cell's volume.

sveticcg [70]3 years ago
4 0

Answer: animal cells have centrosome and lysosomes where plants do not and plant cells have a cell wall where animals do not.

Explanation:

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Setler79 [48]

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Explanation: that’s the answer

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3 years ago
How many protons and electrons are in Hg+
Gala2k [10]

Answer:

There are 80 protons and 80 electrons in Hg+ (Mercury)

Explanation:

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8 0
2 years ago
A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0
yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

 V_s = volume of solution in ml = 150 ml

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles

Now put all the given values in the formula of molality, we get

Molality=\frac{0.178\times 1000}{150}=1.19M

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

1.19\times 15.0=M_2\times 65.0

M_1=0.275M

Therefore, the concentration of KOH for the final solution is 0.275 M

5 0
2 years ago
You wish to construct a buffer of pH=7.0. Which of the following weak acids (w/ corresponding conjugate base) would you select?
Alex17521 [72]

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of  HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C)  pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

D) pKa of HCN = -log(4 x 1 0^-10)=  9.3979

If we consider the  Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution

The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

Hence, HOCl will be chosen for buffer construction.

3 0
3 years ago
What is the value of delta Hrxn for this equation:
Ratling [72]

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

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2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

6 0
2 years ago
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