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3 years ago

Model 1-How is a cell like a factory

Chemistry

1 answer:

anastassius [24]3 years ago

8 0

A cell is like a factory in many ways.

For one, the cell typically contains many organelles that have several varying functions. In a factory, you have many workers who perform different tasks, just as the organelles in the cell do.

Another reason is that cells contain a nucleus or the boss/brain of the cell. In factories, this could be considered the boss of the workplace that tells each and every worker, or organelle, what to do.

The selectively permeable membrane of a cell also resembles a factory as it only lets in workers or special guests. The cell's membrane only lets in specific materials, hence the selectively permeable membrane.

These are just a few ways as to how the cell is like a factory.

Hope this helps!

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Answer: A

Explanation:Idk

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3 years ago

Read 2 more answers

Determine the empirical formulas for compounds with the following percent compositions:

White raven [17]

Answer: a) $CS_2$

b) $CH_2O$

Explanation:

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 15.8g

Mass of S= 84.2 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{15.8g}{12g/mole}=1.32moles$

Moles of S= $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{84.2g}{32g/mole}=2.63moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{1.32}{1.32}=1$

For S = $\frac{2.63}{1.32}=2$

The ratio of C : S = 1:2

Hence the empirical formula is $CS_2$

b) Mass of C= 40 g

Mass of H= 6.7 g

Mass of O = 53.3 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.7g}{1g/mole}=6.7moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.7}{3.33}=2$

For O = $\frac{3.33}{3.33}=1$

The ratio of C : H: O= 1 :2: 1

Hence the empirical formula is $CH_2O$

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3 years ago

4.305 liters to microliters

Hatshy [7]

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1 µL = 10^(-6) L

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3 years ago

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