Answer : The hydroxide ion concentration of a solution is, 
Explanation :
As we know that 
dissociates in water to give hydrogen ion 
and carbonate ion 
.
As, 1 mole of dissociates to give 1 mole of hydrogen ion
Or, 1 M of dissociates to give 1 M of hydrogen ion
So, 0.200 M of dissociates to give 0.200 M of hydrogen ion
Now we have to calculate the hydroxide ion concentration.
As we know that:
![[H^+][OH^-]=1\times 10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D1%5Ctimes%2010%5E%7B-14%7D)
![0.200\times [OH^-]=1\times 10^{-14}](https://tex.z-dn.net/?f=0.200%5Ctimes%20%5BOH%5E-%5D%3D1%5Ctimes%2010%5E%7B-14%7D)
![[OH^-]=5\times 10^{-14}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D5%5Ctimes%2010%5E%7B-14%7D)
Therefore, the hydroxide ion concentration of a solution is,
Answer:
Part A
Kp = 3.4 x 10⁴
Part B
Kp = 2.4 x 10⁻¹⁴
Part C
Kp = 1.2 x 10⁹
Explanation:
2PH₃(g) + As₂(g) ⇌ 2 AsH₃(g) + P₂(g) Kp = 2.9 x 10⁻⁵
Kp = [AsH₃]²[P₂]/[PH₃]²[As] = 2.9 x 10⁻⁵
Part A
it is the inverse of the equilibrium given
Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴
Part B
Is the equilibrium where the coefficients have been multiplied by 3,
Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴
Part C
This is the reverse equilibrium multipled by 2.
Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹
Answer:
Sorry but erm was their supposed to be a image attach with it
Explanation:
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
<h3>
Describe the trends in the atomic size, ionization energy and electronegativity?</h3>
Atomic radius decreases across a period because of nuclear charge increases whereas atomic radius of atoms generally increases from top to bottom within a group because there is again an increase in the positive nuclear charge.
Ionization energy increases when we move from left to right across an period and decreases from top to bottom.
Electronegativity also increases from left to right across a period and decreases from top to bottom.
So we can conclude that atomic size decreases in a period but the ionization energy and electronegativity increases across a period.
Learn more about Electronegativity here: brainly.com/question/24977425
#SPJ1
