g line a) The emission line with the shortest wavelength. b) The absorption line with the shortest wavelength. c) The emission line with the lowest energy. d) The absorption line with the lowest energy. e) The emission line with the lowest frequency. f) The line corresponding to the ionization energy of hydrogen.

a) 4

b) 2

c) 5

d) 2

e) 4

f) 3

Explanation:

The complete question is shown in the image attached to this answer.

We know that the shorter the wavelength, the greater the energy. The greatest energy and shortest wavelength among the emission lines is 4.

Applying the same argument as above, the greatest energy catapults the electron from energy level n=1 to n=4. This will corresponds to the shortest wavelength since energy is inversely proportional to wavelength.  

Frequency is also inversely related to wavelength for two waves travelling at the same speed. Hence, the transition that corresponds to the highest energy and shortest wavelength in the emission spectrum will have the highest frequency.

The emission line with the lowest energy occurs when the electron moves from n=2 to n=1.

The ionization energy is the energy required to remove an electron from the hydrogen atom. It corresponds to the transition n=1 to n=∞

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

fe 55.85 g/mol and h2o 18.02 g/mol.