3. Find out the initial energy level (n) of an electron that results in the emi ssion of light of
wavelength 486 nm in the Balmer series?

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Explanation:

The Balmer series refers to a series of spectral emission lines of the hydrogen atom arising from electronic transitions from any higher level and terminating at the the energy level n= 2.

Using the relation;

1/λ = 1.09737*10^7 (1/2^2 -1/n^2initial)

We now have;

1/486 *10^-9 = 1.09737*10^7 (1/2^2 -1/n^2initial)

0.1875 = 1/2^2 -1/n^2initial

1/n^2initial = 1/2^2 - 0.1875

1/n^2initial = 0.0625

n^2initial = 16

ninitial = 4

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2