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grandymaker [24]
3 years ago
5

Which phenomena support only the wave theory of light?

Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
8 0
Phenomena that supports the only wave theory of light is replfectuon 
Nostrana [21]3 years ago
8 0

Answer:

Diffraction and Interference

Explanation:

The light sometimes acts like a particle and sometimes like a wave, yet the diffraction and interference are proof of the light's wave nature.

First of all, as a definition, the interference is an effect caused by superposition of two systems of waves (from two different sources). (As an example, the interference -distortion- in radio waves)

The diffraction, by the other hand, refers to several events that occur when a wave meets an obstacle. (usually described as a bending of waves around obstacles -likethe water waves- and in other cases as the dissemination of waves, once they passed small openings)

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Most molecules that contain carbon are _____?
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If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
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Answer:

3.64g

Explanation:

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Mass of Cu₂O  = 90.4g

Unknown:

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Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

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Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

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5 0
3 years ago
An exhaled air bubble underwater at 290.
Vinil7 [7]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

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To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

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where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

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From the above equation;

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P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0
3 years ago
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