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Rudiy27
3 years ago
11

Your teacher asked you to identify the stage of mitosis of a specimen under the microscope. You observed that instead of a typic

al round cell shape, the cell has a narrow middle part which almost separates into two bulging ends and which looks like the number 8. What stage is the cell in? *
A. Interphase
B. Anaphase
C. Metaphase
D. Cytokinesis
Chemistry
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

D. Cytokinesis

Explanation:

Mitosis is the cell division that involves the synthesis of two genetically identical daughter cells. Mitotic division occurs in stages namely: prophase, metaphase, anaphase and telophase. Following this four processes is another called CYTOKINESIS

Cytokinesis, which occurs after nuclear division, is the division of the cytoplasm. According to this question, the observation made when viewing the specimen was that "the cell has a narrow middle part which almost separates into two bulging ends and which looks like the number 8". This described the process of CYTOKINESIS in plant cell specifically.

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Water vapor changing to liquid water
kiruha [24]

Explanation:

When water vapour changes to liquid water then this process is known as condensation.

For example, when lid is placed in a hot water filled pan then after sometime vapours appear on the surface of lid. When temperature of water decreases then water vapours convert into liquid form.

Thus, we can conclude that in condensation water vapor changes to liquid water.

4 0
3 years ago
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In the polypeptide Phe-Tyr-Glu-Asp-Ser-Ile-Leu-Ser what is the N-terminal amino acid?
avanturin [10]

Answer:

N-terminal Phe, C-terminal Ser

Explanation:

Amino acids connect like

NH2 -CH(R1) -CO -NH-CH(R2)-CO-.....-NH-CH(Rn)COOH

So, 1st amino acid is N -terminal , and it is Phe.

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7 0
3 years ago
Select True or False from the pull down menus for the following statements.
Shkiper50 [21]

Answer:

First question: 1- False, 2- True, 3- False, 4 -Tue, 5- True.

Second question: 12.7 ºC

Explanation:

First question:

1- When a phase change for a pure substance is taking place under constant pressure, the temperature remains constant, and there's no sensitive heat flowing, but there's latent heat flow, which must be added to separate the molecules and to increase the kinetic energy.

2- When observed the heating and the cooling curve, at the phase change, there is no change in temperature, so it must be a horizontal line, which has a slope equal to 0.

3- Heat is the energy that is transferred by the substances or bodies because of a difference in temperature. The temperature is the measure of average kinetic energy in the molecules, so they are different.

4- As explained above, it's true, that's the definition of temperature.

5- Melting and freezing are the opposite processes and they occur at the same temperature. The difference is that for melting, the substance is absorbing heat, and for freezing it is losing heat, but the heat amount is the same for both process and is calculated by Q = ±m*L, where Q is the heat, m the mass, L the heat capacity, and the signal ± indicates if the substance is absorbing (+) or losing (-) heat.

Second question:

For the conservation of energy, the total amount of heat must be 0. The coin is losing heat, so it must be negative. The water is gaining heat, so it must be positive:

Qw - Qc = 0

Q = m*s*ΔT, where Q is the heat, m is the mass, s is the specif heat, and ΔT the temperature variation (final - initial). Qw is from water and Qc for the coin. The specif heat from the water is 4.184 J/gºC. At the thermal equilibrium, the final temperature must be equal for both.

mw*sw*ΔTw = mc*sc*ΔTc, if the coin is pure silver, sc = 0.233 J/gºC

27.0*4.184*(T - 15.5) = 15.5*0.233*(T - 100)

112.968*(T - 15.5) = 3.6115*(T - 100)

112.968T - 1751.004 = 3.6115T - 361.15

109.3565T = 1389.854

T = 1389.854/109.3565

T = 12.7ºC

So, the final water temperature would be 12.7ºC, which is impossible because it needs to increase. So the coin is not silver pure.

7 0
3 years ago
Please Help!
Whitepunk [10]

We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
</span>

3 0
3 years ago
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Mazyrski [523]
The answer is D i just took the test 
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