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Serhud [2]
3 years ago
7

Determine the new pressure of a sample of hydrogen gas if the volume and moles are constant, the initial pressure is 6.0 atm, an

d it is heated from 15°C to 30.°C. PLEASE HELP ASAP! THANK U SO MUCH!
6.3 atm

5.7 atm

12 atm

3.0 atm
Chemistry
1 answer:
podryga [215]3 years ago
6 0

Answer:

6.3 atm.

Explanation:

Data obtained from the question include:

Initial pressure (P1) = 6 atm

Initial temperature (T1) = 15°C = 15°C + 273 = 288K

Final temperature (T2) = 30°C = 30°C + 273 = 303K

Final pressure (P2) =..?

Since the volume and number of mole of the hydrogen gas sample is constant, the following equation will used to obtain the new pressure:

P1/T1 = P2/T2

6/288 = P2 /303

Cross multiply

288 x P2 = 6 x 303

Divide both side by 288

P2 = (6 x 303) /288

P2 = 6.3 atm

Therefore, the new pressure of the hydrogen gas sample is 6.3 atm.

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what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1
Lostsunrise [7]

Answer:

(<em>n</em> = 7) ⟶ (<em>n</em> = 4)

Explanation:

1. Convert the energy to <em>joules per mole of electrons</em>.  

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2. Convert the energy to <em>joules per electron </em>

<em>E</em> = 55 100/(6.022 × 10²³)

<em>E</em> = 9.150 × 10⁻²⁰ J/electron

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Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes  

\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})

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R_{\text{H}} = the Rydberg constant = 2.178 × 10⁻¹⁸ J

n_{i} and n_{f} are the initial and final energy levels.

9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})      

\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}

n_{f}^{2} = \frac{1 }{ \text{0.062 42}}

n_{f}^{2} = 16.02

n_{f} = \sqrt{16.02}

n_{f} = 4.003 \approx 4

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