Answer:
Think of it this way If you have a phosphorus atom whats its oposites once you found that out you may be able to find the answer
Explanation:
Hope this helps :)
if it is already as low as it can go
Answer:
Condensation
Explanation:
Thermal energy is released in this process
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:
1.2x10⁻⁵M = Concentration of the product released
Explanation:
Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:
A = E*b*C
<em>Where A is the absotbance of the solution: 0.216</em>
<em>E is the extinction coefficient = 18000M⁻¹cm⁻¹</em>
<em>b is patelength = 1cm</em>
<em>C is concentration of the solution</em>
<em />
Replacing:
0.216 = 18000M⁻¹cm⁻¹*1cm*C
<h3>1.2x10⁻⁵M = Concentration of the product released</h3>