<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
Answer:
Metals on the left of the Periodic Table.
Non-Metals on the top-right, plus Hydrogen.
No, darker colors like purple and blue have higher wave frequency which transmit more energy.
Fluorine (F), Chlorine (Cl) and Iodine (I) are all found in the same group on the Periodic Table because they have similar physical properties. Since they are all Halogens, they have 7 valence electrons in their outer shell. In order to get a total of 8, they naturally combine with elements of the same isotope (itself), so D comes close to being correct, but it's not the best answer choice.
As a result, the greatest number of an atom's oxidation state will gradually rise over each period of the periodic table. For instance, the third period's highest value of the oxidation number will fall between 1 and 7.
- The Periodic Table only consistently varies the oxidation numbers of Group 1 and Group 2 metals in their compounds, which are always +1 and +2, respectively.
- Elements have an increasing number of valence electrons that can range from 1 to 8 and move from left to right over time. However, when H or O are added to an element first, the element's valency rises to 4, then falls to zero.
<h3>What causes a rise in the oxidation number?</h3>
An increase in oxidation number results from the loss of negatively charged electrons, whereas a reduction in oxidation number results from the gain of electrons. The result is a rise in the oxidation number of the oxidized element or ion.
<h3>Pattern of the Period 2?</h3>
The trends in Period 2 are significantly more clear-cut. All elements in period 2 experience a decrease in atomic radius, an increase in electronegativity, and an increase in ionization energy as their atomic number rises.
To know more about Periodic table please click here : brainly.com/question/15987580
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