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kicyunya [14]
3 years ago
11

Which model would represent asexual reproduction? explain you reasoning

Chemistry
1 answer:
matrenka [14]3 years ago
8 0

Answer:

Answer)   The first box represent the budding type of asexual reproduction . In budding due to the cell divison at the particular place in the organism , new  individual will produced.

Second picture represent the budding in hydra. Budding is also a asexual reproduction.

Third type of asexual reproduction is found in plants  it is vegetative reproduction.

Fourth is Binary fission . This is also a type of asexual reproduction, in which nucleus dicied into two new daughter cell via mitosis.Explanation:

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4. Write down the differentiate between a. Atom and molecules​
andre [41]

Answer:

Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.

Explanation:

Atom refers to the smallest constituent unit of a chemical element. Molecules refer to a group of two or more atoms that are held together due to chemical bonds.

7 0
3 years ago
Read 2 more answers
You operate a nuclear reactors and want to use fissionable mass that will sustain a slow control what type of mass will you use
puteri [66]

Answer: I believe it is critical mass

Explanation:

3 0
2 years ago
Why is large-scale nuclear fission a rare phenomenon on Earth? A. There isn't enough energy to start a chain reaction in nature.
siniylev [52]

The answer is; C

The radioisotopes in the interior of the earth are not in pure form and have many embedded impurities and are interspersed. Therefore when a neutron is released by a decaying atom, its chances of hitting another radioisotope atom (to continue the chain reaction) are lower than in a manmade nuclear reactor. This way, the radioactive chain reaction in the earth’s interior is not well sustained.


6 0
3 years ago
Read 2 more answers
What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol
statuscvo [17]

Answer:

58.92 g EDTA

Explanation:

315.1 mL = .3151 L    

M = Moles / Liter

.3151 L  x <u>0.5 mol EDTA</u>  x  <u>374 g EDTA</u>  =  58.92 g EDTA

                1 L EDTA             1 mol EDTA

4 0
2 years ago
7. If a waste container has 87.0 mL of 0.234 M acetic acid, how many milliliters of 2.00 M sodium
Naya [18.7K]

Answer:

V_{base}=10.2mL

Explanation:

Hello.

In this case, since the neutralization of the acid requires equal number of moles of both acid and base:

n_{acid}=n_{base}

Whereas we can express it in terms of concentrations and volumes:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we can compute the volume of sodium hydroxide (base) as follows:

V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{87.0mL*0.234M}{2.00M}\\ \\V_{base}=10.2mL

Best regards.

8 0
3 years ago
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