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Ivanshal [37]
2 years ago
5

A complex absorbs photons with an energy of 4.10×10^−19 J. What is the wavelength of these photons? If this is the only place in

the visible spectrum where the complex absorbs light, what color would you expect the complex to be?
Chemistry
1 answer:
MaRussiya [10]2 years ago
6 0

Answer:

483 nm corresponds to blue light hence the complex will appear orange.

Explanation:

Using the formula;

E= hc/λ

Where;

E = energy of the photon

h = Plank's constant (6.6*10^-34Js)

c = Speed of light (3*10^8 ms-1)

λ = wavelength

λ = hc/E

λ = 6.6*10^-34 * 3*10^8/4.10×10^−19

λ = 4.83 * 10^-7 or 483 nm

483 nm corresponds to blue light

Using the colour wheel approach, if a complex absorbs blue light, then it will appear orange.

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For which of the following conversions does the value of the conversion factor depend upon the formula of the substance?
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If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone
GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to  NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

<em>m = 6.46 g of NH3</em>

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3 years ago
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