12.2 C
It has 3 significant figures now.
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:
.
A solid is hard and the molecules are packed together, a liquid can move around freely because the molecules aren't as packed together :)
Answer:
a) Aqueous LiBr = Hydrogen Gas
b) Aqueous AgBr = solid Ag
c) Molten LiBr = solid Li
c) Molten AgBr = Solid Ag
Explanation:
a) Aqueous LiBr
This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.
b) Aqueous AgBr
This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.
c) Molten LiBr
In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.
c) Molten AgBr
The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.
