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3 years ago

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

Chemistry

2 answers:

GalinKa [24]3 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]3 years ago

7 0

I would say C.flagellum,lysosome because I took the test tell me if wrong!

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How many liters of hydrogen (at 0.97atm and 24c) will produce 180 grams of water

slava [35]

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So, 18g water needs 2g H₂

So, 1g water needs 2/18g H₂

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2 years ago

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Taya2010 [7]

Answer:

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2 years ago

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When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal

Mnenie [13.5K]

Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

$V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL$

Next, we compute the densities as shown below:

$\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\ \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL$

In such a way, A is denser is B as it's volume for the same mass is smaller.

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2 years ago

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shtirl [24]

Answer:

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