Is the cell membrane a factory part or worker? Please explain what it does as a factory part or worker

Chemistry

2 answers:

lozanna [386]3 years ago

7 0

Answer:

hey i saw you asked about the kinetic energy lab report on edg and i was wondering if you ever finished it and could send it me plzzzzzzzz

Explanation:

Tema [17]3 years ago

3 0

It is probably a worker bc it lets things in and out of the cell and stops harmful things from entering the cell

You might be interested in

Does temperature change during a chemical reaction?

Anarel [89]

Answer: yes

Explanation:

Not all the time, but temperature change can happen during a chemical reaction.

4 0

3 years ago

Read 2 more answers

Complete each of the definitions with the appropriate phrase. Precision means that measurements are close to Accuracy means that

sergiy2304 [10]

Answer: Precision means that measurements are close to each other . Accuracy means that measurements are close to accepted value

Explanation:

Precision refers to the closeness of two or more measurements to each other.

For Example: If we weigh a given substance five times and you get 5.0 kg each time. Then the measurement is very precise.

Accuracy refers to the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 5.0 kg and one person weighed 4.9 kg and another person weighed 3.9 kg. Then, the weight measured by first person is more accurate.

Thus Precision means that measurements are close to each other . Accuracy means that measurements are close to accepted value.

5 0

2 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$