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
When the first reaction equation is:
AgI(S) ↔ Ag+(Aq) + I-(Aq)
So, the Ksp expression = [Ag+][I-]
∴Ksp = [Ag+][I-] = 8.3 x 10^-17
Then the second reaction equation is:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+
So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2
∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7
by combining the two equations and solve for Ag+:
and by using ICE table:
AgI(aq) + 2NH3 ↔ Ag(NH3)2+ + I-
initial 2.5 0 0
change -2X +X +X
Equ (2.5-2X) X X
so K = [Ag(NH3)2+] [I-] / [NH3]^2
Kf * Ksp = X^2 / (2.5-2X)
8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X
∴ X = 5.9 x 10^-5
∴ the solubility of AgI = X = 5.9 x 10^-5 M
A) Molar mass gold ( Au) = 196.96 g/mol
1 mole Au ----------- 196.96 g
? moles Au ---------- 35.12 g
35.12 x 1 / 196.96
= 0.178 moles of Au
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b) 196.96 g --------------- 6.02x10²³ atoms
35.12 g ---------------- ( atoms ? )
35.12 x ( 6.02x10²³) / 196.96
2.114x10²⁵ / 196.96
= 1.07x10²³ atoms
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