The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the

Answer: The vapor pressure of naphthalene within the flask remains at atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

• The formula employed to determine the enthalpy change for the reaction is:  

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

The provided information includes:

Substituting the values into the previous equation produces:

• The formula utilized to compute Gibbs free energy change is of a reaction:

The equation for the enthalpy change for the reaction is:

The given factors include:

By inserting values from the above equation, we arrive at:

• For the calculation of (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

where,

= Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant =

T = temperature =

= equilibrium constant at 25°C =?

Inserting values into the above equation yields:

• To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

where,

= Equilibrium constant at 35°C =?

= Equilibrium constant at 25°C =

= Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant =

= Initial temperature =

= Final temperature =

By plugging values into the equation above, we obtain:

• In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for , which is:

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for will equal

Consequently, the partial pressure of naphthalene at 35°C is atm.

The double-slit experiment serves as a renowned method to exemplify concepts in quantum mechanics. Specifically, it highlights the idea of wave-particle duality. Employing a light wave shows diffraction and interference, which are typical characteristics of wave behavior. Unexpectedly, using an electron beam produces an interference pattern as well, indicating that electrons can exhibit wave-like properties.

Explanation:

The optical phenomenon would nearly resemble, yet be entirely distinct from, that involved with the exploitation of light. Interference and diffraction are the characteristics distinguishing waves from particles: waves can interfere and disperse, whereas particles cannot.

Light curves around obstacles akin to waves, and this bending results in the single-slit diffraction pattern.

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

• The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
• Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

The molarity of Cu²⁺ is: