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luda_lava [24]
3 years ago
11

On the atomic level energy and matter exhibit the characteristics of

Chemistry
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

  • <em>On the atomic level, energy and matter exhibit the characteristics of </em><u>both waves and particles.</u>

Explanation:

The dual behavior of subatomic particles as waves and particles is known as the wave - particle duality.

And this principle is the basis of the quantum theory.

Such principle is widely applied to the electrons; i.e., the electrons posses wave and particle propeties, which must be understodd as that some of their properties may be explained as if they were particles and others as if they were waves.

For example, from the particle point of view electrons have mass. You can find in internet that the mass of one electron is about 9.1093837015×10⁻³¹ kg or about 1/1836 times the mass of one proton.)

Since, the point of view of the wave characteristics, electrons have wavelength, Louis de Broglie deduced the equation that relates the wavelength and the mass, through this realtion:

  • Wavelength (λ) =  Planck’s constant (h) divided by the product of the mass (m) and the the speed (v) of the particle.

         λ = h / mv

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Phosphorite is a mineral that contains plus other non-phosphorus-containing compounds. What is the maximum amount of that can be
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Phosphorus can be prepared from calcium phosphate by the following reaction:

2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)

Phosphorite is a mineral that contains Ca_3(PO_4)_2 plus other non-phosphorus-containing compounds. What is the maximum amount of P_4 that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% Ca_3(PO_4)_2 by mass? Assume an excess of the other reactants.

Answer: Thus the maximum amount of P_4 that can be produced is 0.345 kg

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Given mass of phosphorite Ca_3(PO_4)_2  = 2.3 kg

As given percentage of phosphorite Ca_3(PO_4)_2 is = \frac{75}{100}\times 2.3kg=1.725kg=1725g

moles=\frac{\text {given mass}}{\text {Molar mass}}

{\text {moles of}Ca_3(PO_4)_2=\frac{1725g}{310g/mol}=5.56moles

2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)

According to stoichiometry:

2 moles of phosphorite gives = 1 mole of P_4

Thus 5.56 moles of phosphorite give= \frac{1}{2}\times 5.56=2.78moles of P_4

Mass of P_4=moles\times {\text {Molar mass}}=2.78mol\times 124g/mol=345g=0.345kg

Thus the maximum amount of P_4 that can be produced is 0.345 kg

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