QUESTIONS 1. Vinegar contains acetic Acid, CH3COOH. Titration of 5.00g of vinegar with 0.100 M
NaOH requires 33.0cmº to reach the equivalence point.
(a) What is the weight percentage of CH3COOH in vinegar?
(b) If the vinegar has a density of 1.005gcm", what is the molarity of CH3COOH in
vinegar?​

389929.827

Explanation:

33806.78° C

(a) 3.96 (w/w) %

(b) 0.663M

Explanation:

The acetic acid, CH₃COOH, reacts with NaOH as follows:

CH₃COOH + NaOH → CH₃COONa + H₂O

Where 1 mole of acetic acid reacts with 1 mole of NaOH

In the equivalence point of the titration, the moles of acetic acid are equal to moles of NaOH added. To solve this question we need to find the moles of NaOH and thus, the mass of acetic acid

Moles NaOH = Moles acetic acid:

33.0cm³ = 0.033L * (0.100moles / L) = 0.0033 moles

Mass acetic acid -Molar mass: 60g/mol-:

0.0033 moles * (60g / mol) = 0.198g

a) Weight percent is 100 times the ratio of the mass of acetic acid (0.198g) in the mass of vinegar (5.00g):

0.198g / 5.00g * 100 = 3.96 (w/w) %

(b) The volume of the vinegar in the 5.00g is:

5.00g * (1mL / 1.005g) * (1L / 1000mL) = 0.00498L

And the molarity is:

0.0033moles / 0.00498L = 0.663M

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

c. electrical charges of ionic bonds.