Period are going left to right across the periodic table
Groups are going up to down on the periodic table
One kilogram is equal to one thousand grams. Further, one gram is equal to 1000 mg. The conversion is as shown below,
(6.285 x 10³ mg) x (1 g / 1000 mg) x (1 kg / 1000 g)
The numerical value of the operation above is 0.006285 kg.
Answer:
See below
Step-by-step explanation:
Ammonium lauryl sulfate has the structural formula CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂OSO₂O⁻ NH₄⁺.
The long nonpolar hydrocarbon chain and the ionic sulfate end group make it a surfactant.
The ionic end tends to dissolve in water, but the nonpolar chain does not. This makes the compound an excellent <em>foaming agent,</em> so it is used in many shampoos and toothpastes.
The molecules form <em>micelles</em> in water, small spherical shapes with the polar heads outside, facing the water, and the nonpolar tails are inside.
They reduce the surface tension or the water so that, when you brush your teeth or shampoo your hair, the air bubbles are stable and do not break.
<span>A fast way to separate the mixture would be to use a magnet. Iron is magnetic while copper is not. The second method would be to visually separate them since iron and copper distinctly different colors.</span>
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
