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VladimirAG [237]
3 years ago
10

How to convert volts to electron volts?

Chemistry
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

How to convert volts to electron-volts

How to convert electrical voltage in volts (V) to energy in electron-volts (eV).

You can calculate electron-volts from volts and elementary charge or coulombs, but you can't convert volts to electron-volts since volt and electron-volt units represent different quantities.

Volts to eV calculation with elementary charge

The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electric charge Q in elementary charge or proton/electron charge (e):

E(eV) = V(V) × Q(e)

The elementary charge is the electric charge of 1 electron with the e symbol.

So

electronvolt = volt × elementary charge

or

eV = V × e

Example

What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 40 electron charges?

E = 20V × 40e = 800eV

Volts to eV calculation with coulombs

The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C) divided by 1.602176565×10-19:

E(eV) = V(V) × Q(C) / 1.602176565×10-19

So

electronvolt = volt × coulomb / 1.602176565×10-19

or

eV = V × C / 1.602176565×10-19

Example

What is the energy in electron-volts that is consumed in an electrical circuit with voltage supply of 20 volts and charge flow of 2 coulombs?

E = 20V × 2C / 1.602176565×10-19 = 2.4966×1020eV

Explanation:

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This is the right answer because in synthesis reactions two or more reactants bound to form only one product.

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This answer is incorrect because this is an example of a decomposition reaction in which one reactant forms two or more products.

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Sodium chloride, NaCl forms in this reaction between sodium and chlorine. 2Na(s) + Cl2(g) → 2NaCl(s) How many moles of NaCl resu
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Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of Cl_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Cl_2=3.9mol

2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)

As Na is the excess reagent, Cl_2 is the limiting reagent as it limits the formation of product.

According to stoichiometry :

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