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<h2><em><u>Technically yes but no because you have to fill the 3s orbital before the 5s orbital</u></em></h2>
<em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em><em><u>✨</u></em>
Answer:The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O2, exhibits strong, broad absorption across the 2500-3200 cm^1 region and an intense absorption at 1715 cm'^-1. Relative absorption intensity: (s)=strong, (m)-medium, (w) weak. What functional class(cs) docs the compound belong to List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly. Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm^1. The functional class(es) of thla compound is(are) alkane (List only if no other functional class applies.) alkene terminal alkyne internal alkyne arene alcohol ether amine aldehyde or ketone carboxylic acid ester nitr
Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
