For hydrogen, what is the wavelength of the photon emitted when an electron drops from a 4d orbital to a 2p orbital in a hydrogen atom? The Rydberg constant is 1.097 × 10-2 nm-1.

486.17 nm

Explanation:

Using the formula   to calculate for the wavelength of the photon emitted. where beta means a substitute for  wavelength (λ) in the above equation.

Our given parameters include ;

The Rydberg constant is 1.097 × 10-2 nm-1.

an electron drops from a 4d orbital to a 2p orbital; which implies the n₂ & n₁ respectively.

Substituting them to the above equation; we can calculate for our wavelength easily;

∴

Since we've earlier said our (β) serves as a substitute for wavelength (λ)

Thus; λ = 0.002056875⁻¹

λ = 486.174 nm

λ ≅ 486.17 nm

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: