Answer:
![K_c\text{ = }\frac{[O_2][H_2]\placeholder{⬚}^2}{[H_2O]\placeholder{⬚}^2}](https://tex.z-dn.net/?f=K_c%5Ctext%7B%20%3D%20%7D%5Cfrac%7B%5BO_2%5D%5BH_2%5D%5Cplaceholder%7B%E2%AC%9A%7D%5E2%7D%7B%5BH_2O%5D%5Cplaceholder%7B%E2%AC%9A%7D%5E2%7D)
Explanation:
Here, we want to write the equilibrium constant expression
To write this, we raise the concentrations of the reactants and products to the coefficient before them. These concentrations are represented by square brackets in which the chemical formula of the compound is placed
We place the representation of the products over that of the reactants
We have the expression written as follows:
![K_c\text{ = }\frac{[H_2]\placeholder{⬚}^2[O_2]}{[H_2O]\placeholder{⬚}^2}](https://tex.z-dn.net/?f=K_c%5Ctext%7B%20%3D%20%20%7D%5Cfrac%7B%5BH_2%5D%5Cplaceholder%7B%E2%AC%9A%7D%5E2%5BO_2%5D%7D%7B%5BH_2O%5D%5Cplaceholder%7B%E2%AC%9A%7D%5E2%7D)
Answer: 1
Explanation: This is true.
Answer:
N = N1V1 +N2V2 +N3V3
(resultant ) ________________
V1+V2+V3
Answer:
The element with electron configuration 1s² 2s² 2p⁶ will most likely not react with an element having seven Valence electrons.
The electron configuration of the element in discuss is 1s² 2s² 2p⁶.
The element has enough electrons to fill it's energy level, n = 2 shell.
In essence, the element in discuss is unreactive as it has attained it's octet configuration and as such is neither in need of an electron nor ready to donate an electron.
As such, although an atom having seven Valence electrons is highly electronegative and as such is an electron attractor, the element with the full octet configuration does not react with it.
This unreactive nature of noble gases is attributed to the full octet configuration of noble gases.
P.S: The electron configuration above is the electron configuration of Neon, Ne.
Let us assume that there is a 100g sample present. The respective mass of each element will then be:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Now, we divide each constituent's mass by its Mr to obtain the moles of each
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Dividing by the smallest number:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula is
C10H12NO
