Potassium oxide: K₂O.
There's no need for prefixes since K₂O is an ionic compound.
<h3>Explanation</h3>
Find the two elements on a periodic table:
- Potassium- K- on the left end of period four.
- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
- Oxygen is a nonmetal.
A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:
- Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
- Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion
when it combines with metals.
The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each
ion is twice that on a
ion. Each
would pair up with two
. Hence the subscript in the formula:
.
There are two classes of compounds:
- Covalent compounds, which need prefixes, and
- Ionic compounds, which need no prefix.
Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide
, the prefix di- indicates that there are two oxygen atoms in the formula
. However, there's no need for prefix in ionic compounds such as
.
<span>The ability of an atom to attract the shared electrons in a covalent bond is its:</span>electronegativity.
Option no. 2 subtracting the atomic number from the mass number
Example: Zn - 65 atomic weight or mass/mass number
30 atomic number
65-30 = 35 number of neutrons
Answer:
435.38 L
Explanation:
From the question given above, the following data were obtained:
Initial mole (n₁) = 3.25 mole
Initial volume (V₁) = 100 L
Final mole (n₂) = 14.15 mole
Final volume (V₂) =?
The final volume occupied by the gas can be obtained as follow:
V₁/n₁ = V₂/n₂
100 / 3.25 = V₂ / 14.15
Cross multiply
3.25 × V₂ = 100 × 14.15
3.25 × V₂ = 1415
Divide both side by 3.25
V₂ = 1415 / 3.25
V₂ = 435.38 L
Thus, the final volume of the gas is 435.38 L