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max2010maxim [7]
3 years ago
5

Using the Rydberg equation, calculate the energy for the following electronic transitions in a hydrogen atom and label each as a

n absorption or an emission. When calculating, watch your signs!
n = 3 → n = 1

a. Energy calculation
b. absorption or emission?
Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

-1.94  * 10^-18 J

Since the electron moved from a higher to a lower energy level (n = 3 → n = 1) it is an emission.

Explanation:

From Rydberg equation;

E = -RH(1/n^2final - 1/n^2initial)

For a transition from  n = 3 → n = 1

RH = 2.18 * 10^-18 J

E = -(2.18 * 10^-18) (1/1^2 - (1/3^2)

E = -(2.18 * 10^-18) (1-1/9)

E= -(2.18 * 10^-18) (8/9)

E = -1.94  * 10^-18 J

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A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
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Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

⇒ P = \frac{m R T}{V}

Put all the values in above formula we get

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Step 2: measure the area of the top of the syringe
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