Answer: The coefficient for the diatomic oxygen (O2) is 3.
Explanation:
To know the coefficient for the diatomic Oxygen, we need to balance the equation.
Fe + O2 -------> Fe2O3
LHS of the equation; Fe = 1 , O2 = 1
RHS of the equation; Fe = 2 , O = 3
∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3
Doing that will give the balance equation which is;
4 Fe + 3 O2 --------> 2 Fe2O3
The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.
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:

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>
A compound is when two or more elements are joined together. The compound that is created with two hydrogens and one oxygen is h2o or water.
Answer:
2 C4H10 + 5 O2 → 4 CH3CO2H + 2 H2O.
Explanation:
Light naphtha components are readily oxidized by oxygen or even air to give peroxides, which decompose to produce acetic acid according to the chemical equation, illustrated with butane .