Answer:
The new volume of the gas is 32L
Explanation:
P1 = 16atm
V1 = 4L
P2 = atm
V2 = ?
According to Boyle's law, the volume of a given mass of gas is inversely proportional to its volume provided the temperature remains constant.
P1 * V1 = P2 * V2
V2 = (P1 * V1) / P2
V2 = (16 * 4) / 2
V2 = 64 / 2
V2 = 32L
The new volume of the gas is 32L
Answer:
Sodium bicarbonate
Explanation:
Sodium bicarbonate ( NaHCO₃ ) -
Sodium bicarbonate , according to the IUPAC nomenclature , its name is sodium hydrogen carbonate ,and in common terms also refereed to as baking soda .
It is a white crystalline solid , it is basic in nature .
<u>The cation and anion of this salt are the sodium ion ( Na⁺) and the anion bicarbonate anion (HCO³⁻) .</u>
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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:
