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andrey2020 [161]
4 years ago
14

Which two structures would provide a positive identification of an animal cell under a microscope?

Chemistry
2 answers:
GalinKa [24]4 years ago
8 0

Answer:

C.flagellum, lysosome

Explanation:

Flagellum and lysosome are the most appropriate structures for providing positive identification of an animal cell under a microscope.

Flagellum are thin filaments, formed by microtubules in their central portion, that allow the cell to swim by liquid means. Usually a flagellate cell has only one flagellum, which can average 200 µm in length and 0.2 µm in diameter. Flagellum are structures located in some specific cell types, which would help to identify a cell.

Lysosomes are organelles present in the cytoplasm of the vast majority of eukaryotic cells, which facilitates their identification. Inside the lysosomes we can find a lot of digestive enzymes.

lisabon 2012 [21]4 years ago
7 0
I would say C.flagellum,lysosome  because I took the test tell me if wrong!
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7. What is the voltage when the resistance is 6 ohms and the current is 8 amps?
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For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of t
Mashcka [7]

Answer:

fraction of vacancies for this metal FV = 1.918*10⁻⁴

Explanation:

Given:

The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )

But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol

Density of the material is given in g/cm3 we need to convert it to g/m³

Density of material ( p ) in g/m³ :

To convert we know that

1 g/cm³ = 1000000 g/m³ then

7.40 * ( 1000000 ) = 7.40*10⁶ g/m³

So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³

Given Atomic mass = 85.5 g/mol

To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above

N = NA * p / A

N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5

N = 4.45*10³⁰ / 85.5

N = 5.212*10²⁸ atoms/m³

We can now Calculate the fraction of vacancies using the formula below;

Fv = Nv / N

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fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴

8 0
3 years ago
Phenol, c6h5oh, is a stronger acid than methanol, ch3oh, even though both contain an o]h bond. Draw the structures of the anions
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The structures of the anions resulting from loss of 1 H from phenol and methanol is shown in the image.

8 0
3 years ago
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Answer:

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Explanation:

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