The volume did not change, it remained at 20 ml
<h3>Further explanation</h3>
Given
20 ml a sample gas at STP(273 K, 1 atm)
T₂=546 K
P₂=2 atm
Required
The volume
Solution
Combined gas Law :

Input the value :

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
Learn more: brainly.com/question/16243729
In all atoms, the number of protons and the number of electrons is always the same. The number of neutrons is very roughly the same as the number of protons, but sometimes it's rather more. The number of protons in an atom is called the atomic number and it tells you what type of atom you have.
Answer:
The correct answer is option D.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

Rate of the reaction:
![R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BCO%5D%7D%7Bdt%7D)
Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.
Given rate expression of the reaction:
![R = k[NO2]^2[CO]](https://tex.z-dn.net/?f=R%20%3D%20k%5BNO2%5D%5E2%5BCO%5D)
Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'
![R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R](https://tex.z-dn.net/?f=R%27%3Dk%282%5Ctimes%20%5BNO_2%5D%29%5E2%282%5Ctimes%20%5BCO%5D%29%3D8%5Ctimes%20k%5BNO2%5D%5E2%5BCO%5D%3D8R)
Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.
Hence, none of the given statements are true.