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3 years ago

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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Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd

harkovskaia [24]

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3 years ago

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Salad (lettuce,tomato,and onion) a pure substance or a mixture ?

balandron [24]

Answer:

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3 years ago

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If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

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3 years ago

Scientific notation for 706000

Sedbober [7]

706000 in scientific notation is 7.06x10^5

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3 years ago

What coefficients must be added to balance the following reaction? _____ Sn + _____ H3PO4 yields _____ H2 + _____ Sn3(PO4)4

Stella [2.4K]

The given unbalanced equation is as follow,

Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄

In above equation first let get start with Sn, on left side its one and on right side its 3. So, multiply Sn on left side with 3 to balance Sn, Hence,

3 Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄

Now, balance Phosphorous, on left side it counts one and on right side it counts 4, so multiply H₃PO₄ on left side with 4 to balance P, Hence,

3 Sn + 4 H₃PO₄ → H₂ + Sn₃(PO₄)₄

Oxygen on both left and right are 16 in number, Hence it is balanced,

Hydrogen on left hand side are 12 in number, while that on right hand side only 2, So, multiply H₂ on right side by 6 to balance hydrogen atoms.

3 Sn + 4 H₃PO₄ → 6 H₂ + Sn₃(PO₄)₄

Above equation is balanced, and coefficients are highlighted bold and underlined.

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3 years ago