Part a the emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. if there are 1.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission? express your answer numerically in joules.

part b during an emission, electrons move from a higher energy orbital to a lower energy orbital. which of the following are valid transitions that produce lines in the emission spectrum of zn?

check all that apply.

1-[ar]4s13d106s1→[ar]4s23d10

2-[ar]4s23d10→[ar]4s23d104p2

3-[ar]4s23d10→[ar]3d10

4-[ar]4s23d10→[ar]4s13d11

5-[ar]3d10→[ar]4s23d10

6-[ar]4s23d10→[ar]4s13d106s1

(a)

(b)

1-

Explanation:

(a)

Where,  

h is Plank's constant having value

c is the speed of light having value

is the wavelength

Given,

n is the number of atoms ,

(b)

Transition from higher level to lower level is : -

1-

Electron from 6s which has higher energy falls to 4s which has lower energy.

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.

its mg hope this