Someone send me a pic of a safety cartoon [drawn] neatly and all it needs is

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

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__

Anastaziya [24]

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

The ratio of atoms in lithium sulfide; Li : S = 2 : 1

Explanation:

In calcium bicarbonate: $Ca(HCO_3)_2$

In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

Number of Oxygen atoms = 3 × 2 = 6

So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide : $Li_2S$

In a molecular formula of lithium sulfide there are:

Number of Lithium atoms = 1 × 2 = 2

Number of Sulfur atoms = 1

So, the Li : S = 2 : 1

8 0

3 years ago

A given atom has 3 protons, 4 neutrons, and 3 electrons. Describe how you can use those three values to describe the structure,

ELEN [110]

Charge # = protons - electons

Mass # = protons + neutrons

so that would be

3-3= charge#
3+4= mass#

3 0

3 years ago

Question 7 of 15

Crazy boy [7]

Answer: 0.4 moles

Explanation:

Given that:

Volume of gas V = 11L

(since 1 liter = 1dm3

11L = 11dm3)

Temperature T = 25°C

Convert Celsius to Kelvin

(25°C + 273 = 298K)

Pressure P = 0.868 atm

Number of moles N = ?

Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)

9.548 atm dm3 = n x 24.47atm dm3mol-1

n = (9.548 atm dm3 / 24.47atm dm3 mol-1)

n = 0.4 moles

Thus, there are 0.4 moles of the gas.

3 0

3 years ago