Answer:
Explanation:
Hello there!
In this case, given the solubilization of cadmium (II) hydroxide:
The solubility product can be set up as follows:
![Ksp=[Cd^{2+}][OH^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCd%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:
Regards!
Answer is: 0.102 moles of HCl would react.
Balanced chemical reaction:
2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).
V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.
V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.
c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.
n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).
n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.
n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.
From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.
n(HCl) = 2 · n(Sr(OH)₂).
n(HCl) = 2 · 0.0051 mol.
n(HCl) = 0.0102 mol; amount of the hydrochloric acid.
Hi!
The correct options would be:
1. Cathode - <em>reduction</em>
The cathode is the negatively charged electrode, and so has an excess of electrons. Cations (positively charged ions) are attracted to the cathode, and gain electrons to acquire a neutral charge. The process in which a gain of electron occurs is called reduction.
2. Anode - <em>oxidation</em>
The opposite occurs at the anode which is positively charged and attracts negatively charged ions, anions. These anions lose their electrons at the anode to acquire a neutral charge, and the process involving loss of electrons is known as oxidation.
3. Salt Bridge - <em>ion transport </em>
Salt bridge is a physical connection between the the anodic and cathodic half cells in an electrochemical cell and is a pathway that facilitates the flow of ions back and forth these half cells. Salt bridge is involved in maintaining a neutral condition in the electrochemical cells, and its absence would result in the accumulation of positive charge in the anodic cell, and negative charge in the cathodic cell.
4. Wire - <em>electron transport </em>
Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.
Hope this helps!
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.
Ca= 40.1
O= 16.0
H= 1.01
keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H
so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
