Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH

Molar mass of water=16+2(1.01)=18.02g
Number of moles of water

Now, mole fraction of NaOH
=

=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
C) the study of the composition of the atom.
Explanation:
A research can be defined as a systematic investigation or careful consideration of study with respect to a particular problem using scientific methods such as collection of data, documenting critical information, analysis of data, and the establishment of facts in order to reach new conclusions.
Similarly, a basic research is an approach to research that's typically theoritical and it's aimed at developing a theory, searching for the truth or gain a better understanding about a phenomenon, subject, or basic laws on nature.
In this context, an example of basic research is the study of the composition of the atom.
An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.
Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.
Answer: -
Acetic acid
Explanation: -
The intermolecular force of attraction depends on the strength or extent of Hydrogen bonding present in a substance.
Benzene and chloroform does not have hydrogen bonding being non polar molecule.
Water has hydrogen bonding being polar.
However acetic acid being most polar has the maximum hydrogen bonding.
Thus acetic acid has the strongest intermolecular forces of attraction.
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9