A 200.0 mL sample of a saturated solution of PbI 2 is given to be allowed to evaporate until dry. The K sp value for PbI 2 , if 240 mg of solid PbI 2 was collected after the completion of the evaporation process is to be calculated. Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, K sp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of A x B y is calculated by the formula, K sp = [ A ] x [ B ] y
A 200.0 mL sample of a saturated solution of PbI 2 is given to be allowed to evaporate until dry. The K sp value for PbI 2 , if 240 mg of solid PbI 2 was collected after the completion of the evaporation process is to be calculated. Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, K sp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of A x B y is calculated by the formula, K sp = [ A ] x [ B ] y
Solution Summary: The author explains that the solubility product, K_sp, is the equilibrium constant that is applied when salt partially dissolves in a solvent.
Interpretation: A
200.0mL sample of a saturated solution of
PbI2 is given to be allowed to evaporate until dry. The
Ksp value for
PbI2, if
240mg of solid
PbI2 was collected after the completion of the evaporation process is to be calculated.
Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product,
Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of
AxBy is calculated by the formula,
Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following:
(a) match structures of isomers given their mass spectra below (spectra A and spectra B)
(b) Draw the fragments given the following prominent peaks from
each spectrum:
Spectra A m/2 =43 and 1/2-57
spectra B m/2 = 43
(c) why is 1/2=57 peak in spectrum A more intense compared
to the same peak in spectrum B.
Relative abundance
Relative abundance
100
A
50
29
29
0
10
-0
-0
100
B
50
720
30
41
43
57
71
4-0
40
50
60 70
m/z
43
57
8-0
m/z = 86
M
90 100
71
m/z = 86
M
-O
0
10 20 30
40 50
60
70
80
-88
m/z
90
100
Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass
spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained.
Draw the structures of these fragments.
For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.