Answer Each Question In No More Than 100 Words

Answer Each Question In No More Than 100 Words Picture Attached For F

1. Molecules capable of forming hydrogen bonds with themselves include water (H₂O), ethanol (C₂H₅OH), and 1-butanol (C₄H₁₀O). These molecules possess highly electronegative atoms like oxygen or nitrogen bonded to hydrogen, facilitating hydrogen bonding. This intermolecular attraction significantly influences their physical properties, such as higher boiling points compared to similar-sized molecules lacking hydrogen bonding capabilities.

2. 1-Butanol has the highest boiling point among the listed compounds due to its ability to form hydrogen bonds, resulting in stronger intermolecular forces. Pentane and diethyl ether have weaker van der Waals and dipole interactions, respectively, while 1-propanol and 2-butanol, though capable of hydrogen bonding, have slightly lower boiling points than 1-butanol because of differences in molecular structure and chain length.

3. To separate pentane and 1-butanol, use distillation exploiting their differing boiling points; pentane (boiling point ≈ 36°C) distills first, leaving 1-butanol (boiling point ≈ 117°C) behind. The process exploits their volatility differences and polar vs. nonpolar nature: pentane’s nonpolar character makes it less soluble in water, aiding separation.

4. Moving the piston downward decreases container volume, increasing ethanol’s vapor pressure and rate of evaporation temporarily. Condensation rate also increases initially due to higher vapor concentration, until equilibrium is reestablished where evaporation and condensation rates are equal, balancing the ethanol amounts in liquid and vapor phases.

5. Increasing volatility order: 3-chloroethane, C₂H₅Cl, (most polar, less volatile)

6. H₂Se > H₂S due to larger size and polarizability; CH₃OH > CH₃SH due to stronger hydrogen bonding; CH₃CH₂OH > CH₃OCH₃ because of higher intermolecular hydrogen bonds; CO₂ > CS₂ because CO₂ has some polarity and quadrupole moments; N₂O > CO₂ since N₂O is a weakly more polar molecule.

7. Hexane and water are immiscible because hexane is nonpolar, relying on London dispersion forces, while water is highly polar with hydrogen bonding. These differing intermolecular interactions prevent mixing despite both molecules attracting each other within their own phases.

8. At 100°C, pKw = -log(51.3 x 10⁻¹⁴) ≈ 13.29. Since pH = -0.5 x pKw, pH ≈ 7 - 0.5 x (13.29 - 7) ≈ 6.36, so water becomes slightly acidic at 100°C but remains approximately neutral for practical purposes.

9. H₂SO₄ > H₂SO₃ due to more sulfur-oxygen bonds and stronger oxidation state; HF > HI because the HF bond is less polarizable but forms stronger hydrogen bonds; CH₂ClCOOH > CH₃COOH because the Cl atom withdraws electron density, stabilizing the conjugate base; H₂O > H₂O₂, as the peroxide’s O–O bond is weaker and less stable, making H₂O more acidic.

Paper For Above instruction

Intermolecular interactions determine the physical and chemical properties of molecules, influencing various phenomena such as boiling points, solubility, and reactivity. Hydrogen bonding, dipole-dipole interactions, van der Waals forces, and ionic interactions are key types of intermolecular forces that vary in strength and impact behavior significantly.

Many molecules, such as water (H₂O), ethanol (C₂H₅OH), and 1-butanol (C₄H₁₀O), are capable of forming hydrogen bonds due to the presence of highly electronegative atoms like oxygen bonded to hydrogen. These hydrogen bonds are responsible for higher boiling points, surface tension, and solubility in water. Molecules like pentane and diethyl ether lack hydrogen bonding; thus, they exhibit lower boiling points due to weaker intermolecular forces.

The boiling points of organic compounds largely depend on the types and strengths of intermolecular forces. For example, 1-butanol exhibits a higher boiling point than pentane because its capacity for hydrogen bonding results in stronger intermolecular attractions. Conversely, pentane relies primarily on London dispersion forces, which are weaker, leading to a lower boiling point, which accounts for the effectiveness of distillation as a separation technique.

Distillation is an effective method for separating pentane from 1-butanol by exploiting their vastly different boiling points. Heating the mixture causes pentane to vaporize first due to its lower boiling point, and it can be condensed in a separate receiver. 1-Butanol remains in the liquid phase until higher temperatures are reached, allowing for efficient separation based on their properties.

When a piston moves down, decreasing the volume of a container with liquid ethanol, the vapor pressure increases initially, leading to a higher rate of evaporation. To restore equilibrium, condensation occurs at a faster rate, balancing the vapor and liquid phases. This dynamic process continues until the rates stabilize, establishing a new equilibrium state.

Volatility order typically correlates with the strength of intermolecular forces. Methane (CH₄) and ethane (C₂H₆) are highly volatile due to weak van der Waals forces. Ethanol (C₂H₆O) has hydrogen bonding, lowering its volatility. Chlorinated ethane (C₂H₅Cl) exhibits intermediate volatility. Thus, the increasing order is C₂H₅Cl

When comparing the boiling points of specific pairs, molecules with stronger intermolecular forces have higher boiling points. H₂Se has a higher boiling point than H₂S due to larger size and polarizability. CH₃OH exhibits a higher boiling point than CH₃SH because of stronger hydrogen bonding. Ethanol (CH₃CH₂OH) has a higher boiling point than dimethyl ether (CH₃OCH₃) owing to hydrogen bonding. CO₂ has a higher boiling point than CS₂ because of polarity and quadrupole moments, and N₂O's slightly higher boiling point results from weak dipole interactions compared to CO₂.

The immiscibility of hexane and water arises from their differing intermolecular interactions: hexane interacts via nonpolar London dispersions, while water’s extensive hydrogen bonding network creates a polar environment. These incompatible forces prevent the mixing of these liquids despite bulk attraction within each phase.

At 25°C, water has a pH of 7.0. Increasing the temperature to 100°C results in an increased ionization constant (Kw), making water slightly more ionized. The pKw at 100°C is approximately 13.29, yielding a pH of around 6.36, indicating that water becomes slightly acidic at higher temperatures, though it remains nearly neutral in terms of acidity.

Higher acidity is observed in H₂SO₄ versus H₂SO₃ due to the former’s stronger oxidizing properties and more complete dissociation. HF exhibits higher acidity than HI because of the high polarity of the H–F bond and its ability to form stronger hydrogen bonds. CH₂ClCOOH is more acidic than CH₃COOH because the chloro substituent stabilizes the conjugate base. H₂O is less acidic than H₂O₂, as the peroxide’s weaker O–O bond facilitates proton donation, making H₂O₂ a stronger acid in comparison.

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