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Ira Lisetskai [31]
2 years ago
6

Consider an atom having four distinct energy levels. If an electron is able to make transitions between any two levels, how many

different wavelengths of radiation could the atom emit?.
Chemistry
1 answer:
Kryger [21]2 years ago
7 0

Energy levels are the electron shells where electrons are found at a fixed distance from the nucleus of the atom. The atom could emit 6 different wavelengths.

<h3>What is wavelength?</h3>

A wavelength is a distance between the adjacent crests in wave signals propagated in a system. Wavelength (\rm \lambda ) is in inverse relation to the frequency of the wave.

When an electron jumps from energy level 1 to 2, 1 to 3, and 1 to 4 one wavelength each is present. Hence, making the total wavelength to be 3, in transition from the first energy level.

Similarly, from energy levels, 2 to 3 and 2 to 4, a total of 2 wavelengths, and from energy levels 3 to 4 one wavelength is produced.

So the total different wavelengths of the radiation that can be emitted will be 3 + 2 + 1 = 6.

Therefore, 6 different wavelengths of radiation will be emitted by the atom.

Learn more about wavelengths here:

brainly.com/question/21419520

#SPJ1

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The question is in the picture below.
pantera1 [17]
The correct option is A.
An oxidation reaction is one in which a substance gives away electrons and becomes oxidized. In the equation given above, the chlorate ion undergoes oxidation reaction and gives away two chlorine ion.
4 0
3 years ago
What is the density of ammonia (NH3) at 293 K and 0.913 atm?
Anastaziya [24]

Answer:  The density of  Ammonia is 0.648 g/l

Explanation:

Density = Mass/ Volume

Mass of one mole of  Ammonia (NH3) = 17.031g

Volume =?

Using the ideal gas law we can determine the volume.

PV = nRT

P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K

Make V the subject of the formular, we then have;

V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm

               V = 24.04358/ 0.913 = 26.3L

Having gotten the value of Volume in this question, we then go back to solve for density.

Density = Mass/ Volume

                17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l

7 0
3 years ago
Given 6193 mL of a gas at 62.3 °C. What is its volume at 38.1 °C?
Luden [163]

Answer:

5746.0 mL.

Explanation:

We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.

V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.

<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>

5 0
3 years ago
Help please and thanks
Likurg_2 [28]
The first one is 2
The second is 1
The third is 6
And the fourth is 3
4 0
3 years ago
Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---&gt; 2HCN (g) + 6H2O (g) delta
AysviL [449]

<u>Answer:</u> The \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

Putting values in above equation, we get:

-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

8 0
3 years ago
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