Complete question:
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.
If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?
Answer:
After 210 s the concentration of SO2Cl2 will be 0.121 M
Explanation:
![ln\frac{[A_t]}{[A_0]} =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D-kt)
where;
At is the concentration of A at a time t
A₀ is the initial concentration of A
k is rate constant = 1.44×10⁻⁴ s⁻¹
t is time
ln(At/A₀) = -( 1.44×10⁻⁴)t
ln(At/0.125) = -( 1.44×10⁻⁴)210
ln(At/0.125) = -0.03024

At/0.125 = 0.9702
At = 0.125*0.9702
At = 0.121 M
Therefore, after 210 s the concentration of SO2Cl2 will be 0.121 M
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, 
The equation is x/25=3.2
*25 *25
-------------
x=80g
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