Answer:
b. mercury
Explanation:
Fishes and some other sea foods are are known to concentrate mercury in their bodies. Consumption of these products gradually accumulate mercury in the body over time.
More specifically, fishes like swordfish, king mackerel and shark are known to concentrate higher amount of mercury than other species of fishes. Hence, limiting their consumption will protect humans from mercury poisoning.
The correct option is b.
Answer:
0.84 mol
Explanation:
Given data:
Moles of ZnCl₂ produced = ?
Mass of Zn = 55.0 g
Solution:
Chemical equation:
2HCl + Zn → ZnCl₂ + H₂
Number of moles of Zn:
Number of moles = mass / molar mass
Number of moles = 55.0 g/ 65.38 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.
Zn : ZnCl₂
1 : 1
0.84 : 0.84
So from 55 g of Zn 0.84 moles of zinc chloride will be produced.
Answer:
... chloride, calcium, potassium, and zinc was signifi- ... of cow and goat milk pasteurization on element retention ... certified American Chemical Society (ACS);. Whatman ... goat milk. Table 2 gives the content of 17 elements of ... found .0026 rag/100 g in raw and .0024 mg/100 ... mg/100 g chloride content (27) and another.
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:
