The final volume up to which 100 .0 mL of 0 .125 M H 2 SO 4 is diluted in order to prepare a 0 .0500 M H 2 SO 4 solution is to be determined. Concept Introduction: The number of moles present in one liter of a solution is known as the molarity of the solution. It is calculated as follows: M = n V Here M is the molarity of the solution, V is the volume of the solution in liters and n is the number of moles of the solute. The process in which the concentration of the solutes in a solution is reduced by adding more solvent to it is known as dilution. The dilution process takes place by adding more amount of solvent to the solution and results in a decrease of the concentration of the solution. But the product of the molarity and the volume should be the same for both the concentrated and the diluted forms. The volume of the diluted solution is calculated by using the following equation: V dil × M dil = V conc × M conc Here, V dil and V conc are the volumes of the dilute and the concentrated solutions, respectively and M dil and M conc are the molarities of the dilute and the concentrated solutions, respectively.
The final volume up to which 100 .0 mL of 0 .125 M H 2 SO 4 is diluted in order to prepare a 0 .0500 M H 2 SO 4 solution is to be determined. Concept Introduction: The number of moles present in one liter of a solution is known as the molarity of the solution. It is calculated as follows: M = n V Here M is the molarity of the solution, V is the volume of the solution in liters and n is the number of moles of the solute. The process in which the concentration of the solutes in a solution is reduced by adding more solvent to it is known as dilution. The dilution process takes place by adding more amount of solvent to the solution and results in a decrease of the concentration of the solution. But the product of the molarity and the volume should be the same for both the concentrated and the diluted forms. The volume of the diluted solution is calculated by using the following equation: V dil × M dil = V conc × M conc Here, V dil and V conc are the volumes of the dilute and the concentrated solutions, respectively and M dil and M conc are the molarities of the dilute and the concentrated solutions, respectively.
Solution Summary: The author explains how the volume of the diluted solution is calculated by using the following equation.
The final volume up to which 100.0 mL of 0.125 M H2SO4 is diluted in order to prepare a 0.0500 M H2SO4 solution is to be determined.
Concept Introduction:
The number of moles present in one liter of a solution is known as the molarity of the solution. It is calculated as follows:
M=nV
Here M is the molarity of the solution, V is the volume of the solution in liters and n is the number of moles of the solute.
The process in which the concentration of the solutes in a solution is reduced by adding more solvent to it is known as dilution.
The dilution process takes place by adding more amount of solvent to the solution and results in a decrease of the concentration of the solution. But the product of the molarity and the volume should be the same for both the concentrated and the diluted forms. The volume of the diluted solution is calculated by using the following equation:
Vdil×Mdil=Vconc×Mconc
Here, Vdil and Vconc are the volumes of the dilute and the concentrated solutions, respectively and Mdil and Mconc are the molarities of the dilute and the concentrated solutions, respectively.
1. Using radii from Resource section 1 (p.901) and Born-Lande equation, calculate the
lattice energy for PbS, which crystallizes in the NaCl structure. Then, use the Born-Haber cycle
to obtain the value of lattice energy for PbS. You will need the following data following data:
AH Pb(g) = 196 kJ/mol; AHƒ PbS = −98 kJ/mol; electron affinities for S(g)→S¯(g) is -201 kJ/mol;
S¯(g)
(g) is 640kJ/mol. Ionization energies for Pb are listed in Resource section 2, p.903.
Remember that enthalpies of formation are calculated beginning with the elements in their
standard states (S8 for sulfur). The formation of S2, AHF: S2 (g) = 535 kJ/mol.
Compare the two values, and explain the difference.
(8 points)
In the answer box, type the number of maximum stereoisomers possible for the
following compound.
A
H
H
COH
OH
=
H
C
Br
H.C
OH
CH
7. Magnesium is found in nature in the form of carbonates and sulfates. One of the major
natural sources of zinc is zinc blende (ZnS). Use relevant concepts of acid-base theory to
explain this combination of cations and anions in these minerals.
(2 points)
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell