How many particles are present in 8.0 moles of silver?

SOVA2 [1]

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:

4 0

3 years ago

2C8H18 (1) + 2502 (g) --> 16CO2 (g) + 18H2O (g)

Makovka662 [10]

Answer:

One gallon of octane produces approximately 7000 L of carbon dioxide.

Note:

I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.

Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.

5 0

2 years ago

Which of these elements could be used in industry to control electron flow?

anzhelika [568]

Answer:

the answer is calcium

Explanation:

6 0

3 years ago

What does a biochemical solar panel have

Viktor [21]

I agree with the person on top have a beautiful day:)

4 0

3 years ago

Urgent help please!!

sveta [45]

Answer:

1. 2.1 moles of Mg

2. 0.72 mole of Mg(OH)2

Explanation:

1. We'll begin by writing the balanced equation for the reaction. This is given below:

3Mg + 2AlBr3 —> 3MgBr2 + 2Al

From the balanced equation above, 3 moles of Mg reacted to produce 2 moles of Al.

Therefore, Xmol of Mg will react to produce 1.4 moles of Al i.e

Xmol of Mg = (3 x 1.4)/2

Xmol of Mg = 2.1 moles.

Therefore, 2.1 moles of Mg is required to 1.4 moles of Al.

2. We'll begin by calculating the number of mole in 26g of water, H2O.

This is illustrated below:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O = 26g

Number of mole of H2O =?

Mole = Mass /Molar Mass

Number of mole of H2O = 26/18

Number of mole of H2O = 1.44 moles

Next, we shall write the balanced equation for the reaction. This is given below:

2HNO3 + Mg(OH)2 —> Mg(NO3)2 + 2H2O

Finally, we can obtain the number of mole of Mg(OH)2 used in the reaction as follow:

From the balanced equation above,

1 mole of Mg(OH)2 reacted to produce 2 mole of H2O.

Therefore, Xmol of Mg(OH)2 will react to produce 1.44 moles of H2O i.e

Xmol of Mg(OH)2 = (1 x 1.44)/2

Xmol of Mg(OH)2 = 0.72 mole.

Therefore, 0.72 mole of Mg(OH)2 was used in the reaction.

3 0

3 years ago