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aivan3 [116]
2 years ago
6

Electron configuration for Ar

Chemistry
1 answer:
timurjin [86]2 years ago
5 0

Answer : The electronic configuration of argon (Ar) is, 1s^22s^22p^63s^23p^6

Explanation :

Element is, Argon (Ar)

Atomic number of argon = 18

The number of electrons = 18

As we know that the number of electrons is equal to the number of protons and the number of proton is equal to the atomic number.

Electronic configuration : It is defined as the arrangement or distribution of electron of atom in a atomic orbital.

Hence, the electronic configuration of argon is,

1s^22s^22p^63s^23p^6

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Iron 3 oxide and carbon react to form iron and carbon dioxide. Balance the equation.
Oksi-84 [34.3K]

Answer:

2 Fe(iii)2O3 + 3 C ==> 2 Fe  + 3 CO2

Explanation:

First of all, you have to translate the words into an equation.

Fe(iii)2O3 + C ==> Fe  + CO2

The easiest way to tackle this is to start with the Oxygens and balance them. They must balance by going to the greatest common factor which is 6. So you multiply the molecule by whatever it takes to get the Oxygens to 6

2 Fe(iii)2O3 + C   ==>     Fe  + 3 CO2

Now work on the irons. There 2 on the left and just 1 on the right. So you need to multiply the iron by 2.

2 Fe(iii)2O3 + C ==> 2 Fe  + 3 CO2

Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3

2 Fe(iii)2O3 + 3 C ==> 2 Fe  + 3 CO2

And you are done.

5 0
3 years ago
What is the answers and pls show work if possible!!
taurus [48]

D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

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46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold


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3 years ago
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However, the gas phases is the phase of matter that perfectly takes the shape of container and occupies all the volume of container as well.

If we recall Dalton's Law of Partial Pressures  we can see that the pressure exerted by the gas components in a container is same like the pressure exerted by the gas alone. These partial pressures of the component of gas combine in such a way that they exert total pressure equal to the constituents' pressure on the container. This way gases occupy all the volume of a container and take the shape of a container they're placed in.

Hope it help!

6 0
3 years ago
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