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Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)


Given s = 7.4×10⁻³ M
So, Ksp is:


Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)

Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
The three type of bridges are beam, arch, and suspension i believe.
Answer:
Rate law: ![k[C_4H_6]^2](https://tex.z-dn.net/?f=k%5BC_4H_6%5D%5E2)
Integrated Rate Law: ![\frac{1}{[C_4H_6]}=\frac{1}{[C_4H_6]_0}+kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BC_4H_6%5D%7D%3D%5Cfrac%7B1%7D%7B%5BC_4H_6%5D_0%7D%2Bkt)

Explanation:
We can see that the graph of time is linear compared to
and the reaction is second order hence we get the rate law from
.
The integrated rate law for second order is
where A is
.
The slope of the graph
w.r.t time is equal to k. The slope of the graph from the table is 0.014 which is equal to k.