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Kryger [21]
3 years ago
15

What is lewis structure?

Chemistry
1 answer:
never [62]3 years ago
6 0

Lewis structures also known as Lewis dot diagrams, Lewis dot formulas, Lewis dot structures,and electron dot structures are diagrams that show the joining together of two things between atoms of a molecule and the <span>alone pairs of electrons that may exist in the molecule.</span>

You might be interested in
I need help with balancing equations I'm doing homework and idk what to do here can you give me the answers please lol
amm1812

It's pretty easy to balance equations! Basically you want to make sure that the number of each compound is equal on both sides of the arrow.

For example number one is

Fe + H2SO4 -> Fe2(SO4)3 + H2

A 3 in front of H2SO4 because there's a subscript of 3 on the right side.

Then a 3 in front of H2 because of the previous step.

Then add a 2 in front of Fe because of the 2 subscript in Fe2(SO4)3

Then add a 1 in front of Fe2(SO4)3 because you already have an equal number of each element.

<u>2</u>Fe + <u>3</u>H2SO4 -> <u>1</u>Fe2(SO4)3 + <u>3</u>H2

I hope this explanation helps! You should really do your homework because practice is everything when it comes to chemistry. You'll need to know how to do it for exams.  

3 0
3 years ago
1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0
3 years ago
Help question 9 is due today also :)
devlian [24]

Answer:

Car 3 with a net force of 12N

Explanation:

The formula is F=MA

Hope this helps friend

8 0
3 years ago
when the concentration of hydrogen ions in a solution is decreased by a factor of ten, the pH of the solution
Vladimir [108]

Answer:

Increases by 1

Explanation:

8 0
3 years ago
How many moles are there in 5.30 x 1024 atoms of silver?
Rus_ich [418]

Answer:

The mass of

4.6

×

10

24

atoms of silver is approximately 820 g.

Explanation:

In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.

1

mole atoms Ag=6.022xx10

23

atoms Ag

Molar mass of Ag =#"107.87 g/mol"#

Multiply the given atoms of silver by

1

mol Ag

6.022

×

23

atoms Ag

. Then multiply times the molar mass of silver.

4.6

×

10

24

atoms Ag

×

1

mol Ag

6.022

×

10

23

atoms Ag

×

107.87

g Ag

1

mol Ag

=

820 g Ag

6 0
2 years ago
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