The volume of one mole of any gas at STP is 22.4 L. So, at STP, the volume of 2.00 moles of hydrogen gas would be (22.4 L/mol)(2 mol H2) = 44.8 L.
Answer:
CRYSTAL
MANY ATOMS THAT ARE ARRANGE IN A REGULAR PATTERN
1:1
Explanation:
Answer:
It could be B or C eighter one
<h2>Complete the table to summarize the properties of the different subatomic particles. </h2>
Explanation:
Atom
It is a smallest particle which cant exist independently.
According To Dalton, atom was indivisible but later on, it was proved that atom can be subdivided into sub atomic particles called electron, proton & neutron.
These subatomic particles have marked properties .
Proton
- It was discovered by E.Goldstein .
- It is positively charged particle
- It is present in nucleus .
- Its mass is equal to 1.6726219 × 10⁻²⁷ kilograms
Neutron
- It was discovered by E.chadwick .
- It is neutral
- It is present inside the nucleus .
- It's mass is equal to 1.674927471×10⁻²⁷ kg
Electron
- It was discovered by J.J Thomson .
- It has negative charge .
- It's mass is equal to 9.10938356 × 10⁻³¹ kilograms
- It is present outside the nucleus in shells .
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:
