Answer:
Ksp = [ Cu+² ] [ OH-] ²
molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol
Ksp = [ Cu+² ] [ OH-] ²
Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰
|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|
|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|
<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|
|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>
Ksp = [ Cu+² ] [ OH-] ²
2.2 ×10-²⁰ = (S)(2S)²= 4S³
![s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}](https://tex.z-dn.net/?f=s%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B2.2%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2020%7D%20%7D%7B4%7D%20%7D%20%20%3D%201.8%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%207%7D%20)
S = 1.8 × 10-⁷ M
The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M
Solubility of Cu (OH)2 =

<h3>
Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>
I hope I helped you^_^
The element with the lowest ionization energy is CESIUM, CS.
Ionization energy is the energy required to remove the most loosely bound electron in an atom of an element. The higher the number of shells in an atom of an element, the lower the ionization energy that will be required to remove the valence electron.
Answer:
Explanation:
Convergent boundaries: where two plates are colliding. Subduction zones occur when one or both of the tectonic plates are composed of oceanic crust. ...
Divergent boundaries – where two plates are moving apart. ...
Transform boundaries – where plates slide passed each other.
Mechanism for oxidation of alkene by KMnO₄ is provided in the attached image:
Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin