They have same chemical properties
for example the elements of group VIII are called noble gases .They are unreactive, odorless and colorless.
C. Ca and Br because they're metals and nonmetals
According to Arrhenius theory of acid and base, Acids are those substances which when dissolved in water produces protons, while, Bases are those substances which when dissolved in water produces Hydroxyl Ions.
Example of Arrhenius Bases:
NaOH ₍s₎ → Na⁺ ₍aq₎ + ⁻OH ₍aq₎
LiOH ₍s₎ → Li⁺ ₍aq₎ + ⁻OH ₍aq₎
Result:
The only negative ion produced in water when Arrhenius Base is dissolved is ⁻OH (Hydroxyl Ion).
Answer:
Elements that fall between those on the left and right sides of the periodic table
Explanation:
Transition metals:
These are present at the center of periodic table.
These are d-block elements.
They include the elements of group 3 to 12 in periodic table.
They have large charge to radius ratio.
They mostly form paramagnetic compounds.
They shoes more than one oxidation state.
They form colored compounds.
They all have high melting and boiling point.
They have high densities.
They form stable complexes.
The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.
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:
