Answer:
For n=3 and l=1=p
It is 3p-orbital.
Magnetic quantum number m
l
have values from -l to +l and total of 2l+1 values.
Forl=1, m
l
values are:
m
l
=−1,0,1 for l=1; total m
l
values =3= Number of orbitals
Each orbital can occupy maximum of two electron
Number of electrons =2×3=6
Thus 6 electrons will show same quantum number values of n=3 and l=1.
Number of elements with last electron in 3p orbitals = 6
Answer:
Equilibrium concentrations of the gases are



Explanation:
We are given that for the equilibrium

Temperature, 
Initial concentration of



We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of



At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values



By solving we get

Now, equilibrium concentration of gases



Answer:

Explanation:
We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.
We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

Flip the ratio so the units of atoms of fluorine cancel each other out.


Condense into 1 fraction.

Divide.

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>
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