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Ira Lisetskai [31]
3 years ago
6

The emission spectrum of hydrogen shows discrete, bright, colored lines. Which characteristic of the Bohr model is best supporte

d by this observation?
Chemistry
2 answers:
natali 33 [55]3 years ago
7 0
The characteristic of the Bohr model that would best support his observation is this assumption: "The energy of the electron in an orbit is proportional to its distance from the nucleus. The further the electron is from the nucleus, the more energy it has." The discrete, bright, colored lines might represent the electrons and its distance from the nucleus. The lights are caused by the energy it has.
Whitepunk [10]3 years ago
3 0
<h2>Answer : Electrons cannot exist in locations other than in specific orbits.</h2><h3>Explanation : </h3>

The emission spectrum of hydrogen shows discrete, bright, colored lines. The characteristic that the Bohr model supported with observation is that electrons cannot exist in location other than in specific orbits.

According to characteristics of Bohr's atomic model;

  • The energy levels of electrons are discrete, this is the reason which is shown by hydrogen emission spectrum as discrete, bright, colored lines.
  • Also, electrons are seen to orbit around the nucleus in orbits which has a set size and energy with a defined path.
  • Electrons can jump from one energy level to another, which results after absorbing or emitting an electromagnetic radiation with a frequency ν.

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A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
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Answer:

The molarity of this solution is 0,09254M

Explanation:

The concentration of the Ni²⁺ solution is:

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0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>

25,00 mL of this solution contain:

0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺

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0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺

Thus, moles of Ni²⁺ that react with CN⁻ are:

4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺

For the reaction:

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Four moles of CN⁻ react with 1 mole of Ni²⁺:

2,9450x10⁻⁴ moles of Ni²⁺ × \frac{4 mol CN^-}{1 molNi^{2+}} = <em>1,178x10⁻³ moles of CN⁻</em>

As the volume of cyanide solution is 12,73mL. The molarity of this solution is:

<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>

I hope it helps!

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