This is an incomplete question, here is a complete question.
A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸
Answer : The concentration of 
ions at equilibrium is, 
Explanation : Given,
Moles of 
= 0.130 mol
Volume of solution = 1 L
Concentration of = Concentration of = 0.130 M
Concentration of 
= 1.20 M
The equilibrium reaction will be:
![Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq%29%2B6NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D)
Initial conc. 0.130 1.20 0
At eqm. x [1.20-6(0.130)] 0.130
= 0.42
The expression for equilibrium constant is:
![K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B%5BNi%28NH_3%29_6%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5BNH_3%5D%5E6%7D)
Now put all the given values in this expression, we get:
Thus, the concentration of ions at equilibrium is,
B) 8
Neon is a noble gas which means it is in the 18th group which has 8 valence electrons. Thus Neon has 8 valence electrons.
Answer:
0.50 mol
Explanation:
The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.
Initially, there are 8.0 moles.
- After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.
- After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.
- After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.
- After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
Answer: Trees, like all plants, are alive and require nutrients to survive. ... This dormancy is what allows trees to survive the cold winter. During dormancy, a tree's metabolism, energy consumption, and growth all slow down significantly in order to endure the harsh season of winter when water and sunlight are more scarce
Explanation: mark me brainlest plz if this helped only !
