Answer:
I'm sure it's option A. Cast Fossil
Your limiting is CuCI2 and the excess is KI (from what i’ve heard from my tc to find it just use the moles or look at the grams)Do you want me to do the qn and give u the ans or?
Explanation:You have more grams of KI than CuCI2
irl example : I need 200g of flour to bake 1 muffin and 100g of butter.But I have 300g of butter and only 200g of flour.This means I can only bake up to 1 muffin since I got excess grams of butter.But to use up all my 300g of butter I need 400g more of flour.Making my butter the excess while my flour the limiting since I have less of it and it also determines how much muffin would I get at the end of the bake.
im sorry if that example sounds clowny T-T
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:

Answer:
The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein
Explanation:
The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:
Initial volume * initial concentration = final volume * final concentration
0.5 mL * 10M = 5mL * final concentration
1.1 M = final concentration = main solution concentration
Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.
Dilution factor = Main solution concentration/tube 4 concentration
Dilution factor = 1.1/0.0088 = 125
I hope my answer helps you
Answer:
this answer is b no u can it very easy
Explanation:
because solid and liquid can compress easily
plz support me