Answer:
Group 18th
Explanation:
The group 18th of periodic table which is composed of Noble gases mainly was missing from the Mendeleev's periodic table. The reason for this is that Noble gases are least reactive, therefore, it was difficult to predict such elements which do not react with other elements.
Answer:
Sodium - malleable, soft, and shiny
Silicon - has properties of both metals and nonmetals
Bromine - highly reactive gas
Argon - non-reactive gas
Explanation:
Sodium is an alkaline metal. Just like other alkaline metals, it's malleable, soft, and shiny.
Silicon is a metalloid. Metalloids are elements that have properties of both metals and nonmetals.
Bromine a highly reactive chemical element. It is a fuming red-brown liquid at room temperature that evaporates to form a similarly coloured gas.
Argon is a noble gas. Just like other noble gases, it's non-reactive.
Answer:
Complete ionic:
.
Net ionic:
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water:
,
, and
. These three salts will exist as ions:
- Each
formula unit will exist as one
ion and one
ion. - Each
formula unit will exist as one
ion and two
ions (note the subscript in the formula
.) - Each
formula unit will exist as one
and two
ions.
On the other hand,
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite
,
, and
(three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each
formula unit will exist as only one
ion and one
ion. However, because the coefficient of
in the original equation is two,
alone should correspond to two
ions and two
ions.
Do not rewrite the salt
because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of
and two units of
. Doing so will give:
.
Simplify the coefficients:
.