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ahrayia [7]
3 years ago
14

Using the particle-in-the-box model for the hydrogen atom and treating the atom as an electron in a one-dimensional box of lengt

h 150. pm, predict the wavelength of radiation emitted when the electron falls from the level with n 3 to that with n 2

Chemistry
1 answer:
Levart [38]3 years ago
8 0
The model has resulted into a formula called the Rydberg formula. It also came about from the electron transfer within the hydrogen atom. The formula is shown in the picture, where RH is the Rydberg constant and λ is the wavelength . Thus, the solution is as follows:

1/λ = (1.0973731×10⁷ m⁻¹)(1/2² - 1/3²)
Solving for λ,
<em>λ = 6.56×10⁻⁷ m</em>


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Is 2AI + 3O2 = 2AI2O3 balanced or unbalanced
Rudik [331]
Balanced there are 2 Al, 6 oxygens on both sides.
7 0
3 years ago
When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo
Georgia [21]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

3 0
3 years ago
PLEASE HELP ME WILL DO ANYTHING ‼️‼️
deff fn [24]

Answer:

In first shell 2 electrons are present and 7 electrons are present in last shell

Explanation:

6 0
3 years ago
Michelle is trying to find the average atomic mass of a sample of an unknown
GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

  • Abundance of isotope A (A%) = 59.34%
  • Mass of isotope A = 113.6459 amu
  • Mass of isotope B = 115.8488 amu
  • Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
  • Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

Average \: atomic \: mass \:  =  \frac{mass \: of \: A \times A\%}{100}  + \frac{mass \: of \: B \times B\%}{100}  \\  \\ Average \: atomic \: mass \:  =  \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\  \\ Average \: atomic \: mass \:  = 114.54 \: amu  \\  \\

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0
2 years ago
If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?
zavuch27 [327]
<span>Feb 19, 2014 - The units of k tell you that this is a second order reaction. So, to solve this, you need to use the integrated rate law for a 2nd order reaction: 1/[A] = kt + 1/[A]o 1/[A] = 0.540/Ms (835 s) + 1/0.00640 1/[A] = 607 [A] = 1.65X10^-3 M.</span><span>
</span>
7 0
3 years ago
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