Answer:
2.86mol/L
Explanation:
Given parameters:
Number of moles of MgCl₂ = 7.15moles
Volume of solution = 2.50L
Unknown:
Molarity of the MgCl₂ solution = ?
Solution:
The molarity of a solution is the number of moles of solute found in a given volume.
Molarity = 
Insert the parameters and solve;
Molarity = 
= 2.86mol/L
Answer:
True
Explanation:
The arrhenius theory describes an acid as a substance that produces excess hydrogen ions when it interacts with water. Therefore, they increase the concentration of hydrogen ions in solutions.
An arrhenius base interacts with water to yield excess hydroxide ions in aqeous solutions.
Answer: four hundred and 18
Explanation:
Simple math showed that only 16 words are possible from a two-letter combination, but a three-letter code produces 64 words. Operating on the principle that the simplest solution is often correct, researchers assumed a three-letter code called a codon.
<u>Explanation:</u>
It is given that polar solutes can be dissolved in polar solvents and non-polar solutes can be dissolved in non-polar solvent.
Alcohol being polar, does not dissolves ionic salt in it.
is a non-polar solvent.
From the given options:
1. NaCl: This is an ionic salt and hence, it will be soluble in water only.
2. 
: Iodine gas is a non-polar solute and hence, will be dissolved in non-polar solvent which is
3. Ethanol: As, it is a polar molecule and is not an ionic salt, therefore it can be soluble in both water and alcohol.
4. Benzene: It is a non-polar molecule and hence, it will be dissolved in
5. 
: Bromine gas is a non-polar solute and hence, will be dissolved in
6. 
: This is an ionic salt and hence, it will be soluble in water only.
7. Toluene: This is a non-polar solute and hence, will be dissolved in
8. 
: This is an ionic salt and hence, it will be soluble in water only.
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0
