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:
a)the ionic radius of phosphorus
Explanation:
phosphorus=2.8.6
for phosphorus to become stable(2.8.8) it gains two electrons thus the ionic radious increases bcz of the gained electrons
Answer:
+60.54 kJ/mol
Explanation:
The enthalpy of formation of a compound is the sum of the enthalpies of the formation of its constituents and the bond of them. To form a compound, energy must be lost, so, they're negative.
For SF4, the enthalpy is formed by the energy of one S, two F, and 4 S-F bond (Hb):
H = - (278.8 + 4*79.0 + 4*Hb)
-775 = -(594.8 + 4Hb)
594.8 + 4Hb = 775
4Hb = 180.2
Hb = +45.05 kJ/mol
For SF6, the enthalpy is formed by the energy of one S, six F, and 6 S--F bonds (Hb):
H = -(278.8 + 6*79.0 + 6*Hb)
-1209 = -(752.8 + 6Hb)
752.8 + 6Hb = 1209
6Hb = 456.2
Hb = +76.03 kJ/mol
Thus, the energy of S--F bond must be the average of these two:
(45.05 + 76.03)/2 = +60.54 kJ/mol
Answer:
Original price: $
54.29
Discount percentage:
20
%
Explanation:
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Answer:
Kinetic energy has a direct relationship with mass, meaning that as mass increases so does the Kinetic Energy of an object. ... Objects with greater mass can have more kinetic energy even if they are moving more slowly, and objects moving at much greater speeds can have more kinetic energy even if they have less mass
