Answer:
The following three isomeric structure are given below.
Explanation:
Structure of the following three isomeric esters with chemical formula C₇H₁₂O₂
Ester #1: methyl 1-methylcyclobutanecarboxylate
Ester #2: (E)-methyl 3-methyl-3-pentenoate
Ester #3: isopropyl 2-methylpropenoate
Answer:
Compound 1 is molecular
Compound 2 is ionic
Compound 3 can't really be decided
Explanation:
A molecular substance does not conduct electricity, has very low melting and boiling points and is held together by very weak intermolecular forces.
An ionic substance conducts electricity in solution or in molten state but never in the pure solid state, has a high melting and boiling point and has a dull appearance most times.
Compounds 1 shows the properties of molecular substances hence it are designated as such.
On the other hand, compound 2 shows the properties of an ionic substance and is also designated as such.
We can't really decide on compound 3 because it shows some properties of ionic substances and some properties of molecular substances.
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
Answer: But-2-enoic acid has
11 Sigma Bonds and
2 Pi Bonds.
Explanation: The sigma bonds which are formed due to head to head overlap of partally filled orbitals are shown in
red color, while Pi bonds which are formed after the formation of sigma bond by overlap of orbitals perpendicular to the sigma bond are shown in
blue color.
