Answer:
pOH= 14.248
[H+]=1.77 M
[OH-]=5.65 x10^-15M
Explanation:
pH+pOH= 14
pOH= 14-pH
pOH=14-(-0.248)
pOH= 14.248
[H+]=10^-pH= 10^-(-0.248)=1.77 M
[OH-]=10^-pOH= 10^-14.248=5.65 x10^-15M
Answer:
<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>
Explanation:
In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.
NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.
Answer:
A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode
Explanation:
Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it. The resulting changes at the anode and cathode are given by the equation:
cathode: Cu²⁺ + 2e⁻ ⇒ Cu
anode: Cu ⇒ Cu²⁺ + 2e⁻
Answer:
Explanation:
A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.
This is an arrhenius Base
According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :
- Sodium hydroxide NaOH
- Potassium hydroxide KOH
A substance that produces an excess of hydrogen ion (H+) in aqueous solution
This is an arrhenius Acid
An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.
Examples are;
- Hydrochloric acid HCl
- Hydroiodic acid HI
- Hydrobromic acid HBr
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble the data in one place.
2Mg + O₂ ⟶ 2MgO
n/mol: 2 5
Calculate the moles of MgO we can obtain from each reactant.
From Mg:
The molar ratio of MgO:Mg is 2:2
From O₂:
The molar ratio of MgO:O₂ is 2:1.
