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:
.
Answer:
The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%
Explanation:
When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.
A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.
In this case:
- mass of solute: 3.55 g
- mass of solution: 3.55 g + 88 g= 91.55 g
Replacing:

Percent by mass= 3.88%
<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
1016)(2.2<span> × </span>1010<span>) = A × </span>10B<span> A= B= Chemistry.</span>