PH+pOH=14
4.5+pOH=14
pOH=14-4.5
pOH = 9.5
We know that
[H⁺][OH⁻]=10⁻¹⁴,so
-log([H⁺][OH⁻])=-log10⁻¹⁴
-log [H⁺]+ (- log[OH⁻])= - log10⁻¹⁴,
- log[H⁺] = pH
- log[OH⁻] = pOH
- log10⁻¹⁴=-(-14)log10=14*1=14
pH + pOH =14
Your answer would be CCI4. CCI4 is bonded by sharing electrons, which means it is covalent bonding
A. add a catalyst is the correct answer.
A catalyst never takes part in the reaction, so it does not affect the reaction.
Hope this helps~
Answer:
their are a it is in the chemicle
Explanation:18 carbon and 4 hydrogen
so it is a toatle of 22 atoms
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>