The molarity of a solution that is prepared by dissolving 15.5 g of NaOH in 1.50 L of solution is
How is molarity of solution calculated?The given values are,
volume = 1.50 L
weight of NaOH = 15.5 g
Mass NaOH= 40 g/mol
Therefore, moles of NaOH is = weight / mass
= 15.5 / 40
= 0.3875
We know,
Molarity= moles of NaOH / volume
= 0.3875 / 1.50
= 0.25833 M NaOH
Therefore, we get the molarity of this solution as 0.25833 M NaOH
What is molarity of a solution?Molarity is defined as the number of moles of solute dissolved in one liter of solution. Also called molarity and represented by the letter "M". Molarity formula M = n/v where M is the molarity n is the number of moles v is the total volume of the solution in liters.
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Match each label below with the appropriate term. Note: there may be more than one correct answer.
A Lewis diagram: two atoms of upper C l connected by a single bond. Both atoms have two electrons above, below, and to the outside of the pair. Arrow a points to the bond; arrow b to the electron dots below the left atom. A second diagram shows upper O single bonded to upper H to the left and below, with two electron dots above and to the right. Arrows c point to each of the bonds.
a
nonbonding electrons
sigma bond
represents two electrons
b
nonbonding electrons
sigma bond
represents core electrons
c
nonbonding electrons
bonding electrons
sigma bond
The Matchup of each label with the appropriate term is given below:
A: Sigma bond & Represents two electrons
B: Nonbonding electrons
C: Bonding electrons & Sigma bond
What is Sigma bond?Sigma bonds, often known as bonds, are the strongest kind of covalent chemical bonding in chemistry. They are created by atomic orbitals directly overlapping one another. Using symmetry group terminology and techniques, sigma bonding for diatomic molecules is most easily defined.
Note that an electron that is not a part of chemical bonding is known as a non-bonding electron. The electron may be restricted to one atom in a lone pair. has the electron distributed throughout the molecule in a non-bonding orbital.
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Answer:B,C
Explanation: