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astraxan [27]
3 years ago
15

You measure the emission spectrum of an atom that you suspect to be hydrogen. You have no absolute calibration of the wavelength

, but the ratios of the first 3 wavelengths in the series that you measure are 1 : 0.741 : 0.661. Which series is this (e.g. Lyman, Balmer, Paschen, Brackett)? Justify your answer.
Chemistry
1 answer:
trapecia [35]3 years ago
5 0

Answer:

Explanation:

What we need to do here is to determine the ratios by using  the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing  energy levels.

1/λ = Rh x ( 1/n₁² - 1/n₂²)

Lyman series  ( n₁=1 and n₂= 2,3 etc)  for the first two lines, the ratios will be:

1/λ₁ /1/λ₂  =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74  (the first ratio)

For Balmer series n₁ = 2  and n₂ = 3,4,5, etc

1/λ₁ /1/λ₂  =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)

Lets use the third line to check our answer:

1/λ₁ /1/λ₂  =(1/4 -1/3²) / (1/4 -1/5²) = 0.66

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

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3 years ago
I really need help on this question.
natita [175]
I need that answer too
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What type of chemical reaction involves 2 or more elements combining to form one?
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Hey there!

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Answer:

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Explanation:

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