In an experiment, two unknown compounds (one an alcohol and the other an ether) of equal molecular mass were dissolved in water.
1 answer:
Answer: B. It is an ether because it is unable to to form a hydrogen bond, so it is less soluble than water
Explanation: Alcohols are more soluble in water as they can form hydrogen bonding with water whereas ethers
are less soluble as they do not form hydrogen bonds with water.
For formation of hydrogen bond, the electronegative atom (F, Oand N) must be attached to hydrogen and in ethers (ROR), there is no hydrogen directly attached to electronegative oxygen atom, thus are less soluble in water.
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In general chemistry, isomers are two or more elements that have the same number of protons but differ in mass number. In organic chemistry, the compounds are cis or trans isomers if they have the same chemical formula, but differ in the placement of functional groups based on molecular geometry. Cis isomer is when two like functional groups are on the same side of the molecules, while trans isomer is when the like functional groups are on opposite sides.
The cis-trans isomers are shown in the picture. As you can see, in the cis isomer, the methane functional group are both in the same side. Same as well with the hydrogen atoms. On the other hand, these functional groups are opposite to each other in the trans isomer.
The cis-trans isomers are shown in the picture. As you can see, in the cis isomer, the methane functional group are both in the same side. Same as well with the hydrogen atoms. On the other hand, these functional groups are opposite to each other in the trans isomer.
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
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