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^_^
It is given that the price of the carpet is $ 28.99 per square yard
Step 1: Convert square yard to square meter
1 square yard = 0.836 square meter
Step 2: Calculate the cost/square meter
Now based on the relation in step (1) we can write
0.836 square meter of carpet costs $ 28.99
Therefore the cost of 1 square meter of the carpet would be
= $ 28.99 * 1 sq meter/0.836 sq meter
= $ 34.68
The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2