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lesya [120]
3 years ago
6

Many scientist compare the parts of a cell to the parts of a factory. Do you think this comparison is fair and useful?

Chemistry
1 answer:
Fittoniya [83]3 years ago
4 0
Yes. Parts of a cell work together just like stations in a factory.
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Why is ozone an important form of oxygen?
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Ozone is made up of 3 atoms of Oxygen and present in stratosphere of our atmosphere. It prevents us from "Ultraviolet rays" of Sunlight so it is an important form of oxygen or us

Hope this helps!
4 0
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Which of the following is not a weakness of the Rutherford model of the atom?
Harrizon [31]
The answer to this is t<span>he atom is mostly empty space.</span>
7 0
3 years ago
Which of these changes will increase the vapor pressure of water in a sealed container?
Vlad1618 [11]

Answer:

Option B

Explanation:

Salt is a non-volatile solute and hence adding salt will increase the boiling point of water and hence reduce the vapor pressure. While on the other hand, adding more water will require more time to boil and hence produce vapor and thus the vapor pressure. Shaking will also not help in increasing the vapor pressure. Thus, only increasing the temperature of the water will create more vapors at a faster rate and hence increase the vapor pressure.

Thus, option B is the correct answer

3 0
2 years ago
Help plz i need it my mom is mad at me for not knowing this
ohaa [14]
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5 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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