1)Identify the atoms that are participating in a covalent bond.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.
A nitrogen atoms can make 3 covalent bonds because it has three unpaired electrons
Answer: Balanced molecular equation :

Total ionic equation:
The net ionic equation:

Explanation:
Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.
The balanced molecular equation will be,

The total ionic equation in separated aqueous solution will be,

In this equation, and are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,

Answer:
b. 3.66x10²³ atoms of chromium.
Explanation:
First we calculate how many moles are there in 31 grams of chromium, using its molar mass:
- Molar Mass of Chromium = 51 g/mol (This can be found on any periodic table)
- 31 g ÷ 51 g/mol = 0.608 mol
Then we <u>calculate how many atoms are there in 0.608 moles</u>, using <em>Avogadro's number</em>:
- 0.608 mol * 6.023x10²³ atoms/mol = 3.66x10²³ atoms
The correct answer is thus option b. 3.66x10²³ atoms of chromium.