A solution is to be defined. The dissolution of an ionic solute such as NaCl in water to form a solution is to be described. The reason for, the strong bonding forces are overcome in a crystal of ionic solute during the formation of a solution,, is to be explained. The reason for the ions in a solution not attract each other so strongly as to reconstitute the ionic solute is to be explained. The reason for a molecular solid such as sugar dissolve in water is to be explained. The forces between water molecules and the molecules of a molecular solid that help the solute dissolve is to be describe. The reason for some substances not dissolve in water at all is to be explained. Concept Introduction: A solution is made up of mainly two components one is solute and other is solution. Any solute and solvent will form a solution is depend on the “like dissolve like” principle. The polar solute only dissolve in polar solvent and a non-polar solute only dissolve in a non-polar solvent.
A solution is to be defined. The dissolution of an ionic solute such as NaCl in water to form a solution is to be described. The reason for, the strong bonding forces are overcome in a crystal of ionic solute during the formation of a solution,, is to be explained. The reason for the ions in a solution not attract each other so strongly as to reconstitute the ionic solute is to be explained. The reason for a molecular solid such as sugar dissolve in water is to be explained. The forces between water molecules and the molecules of a molecular solid that help the solute dissolve is to be describe. The reason for some substances not dissolve in water at all is to be explained. Concept Introduction: A solution is made up of mainly two components one is solute and other is solution. Any solute and solvent will form a solution is depend on the “like dissolve like” principle. The polar solute only dissolve in polar solvent and a non-polar solute only dissolve in a non-polar solvent.
Solution Summary: The author explains that a solution is made up of mainly two components one is solute and another is solvent.
A solution is to be defined. The dissolution of an ionic solute such as NaCl in water to form a solution is to be described. The reason for, the strong bonding forces are overcome in a crystal of ionic solute during the formation of a solution,, is to be explained. The reason for the ions in a solution not attract each other so strongly as to reconstitute the ionic solute is to be explained. The reason for a molecular solid such as sugar dissolve in water is to be explained. The forces between water molecules and the molecules of a molecular solid that help the solute dissolve is to be describe. The reason for some substances not dissolve in water at all is to be explained.
Concept Introduction:
A solution is made up of mainly two components one is solute and other is solution. Any solute and solvent will form a solution is depend on the “like dissolve like” principle. The polar solute only dissolve in polar solvent and a non-polar solute only dissolve in a non-polar solvent.
Using the Nernst equation to calculate nonstandard cell voltage
A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction:
MnO2 (s)+4H* (aq)+2Cr²+ (aq) → Mn²+ (aq)+2H₂O (1)+2Cr³+ (aq)
+
2+
2+
3+
Suppose the cell is prepared with 7.44 M H* and 0.485 M Cr²+ in one half-cell and 7.92 M Mn² and 3.73 M Cr³+ in the other.
Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
☐
x10
μ
Х
5
?
000
日。
Calculating standard reaction free energy from standard reduction...
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG° for the following redox reaction.
Be sure your answer has the correct number of significant digits.
NO (g) +H₂O (1) + Cu²+ (aq) → HNO₂ (aq) +H* (aq)+Cu* (aq)
kJ
-
☐ x10
x10
olo
18
Ar
Calculating the pH of a weak base titrated with a strong acid
b
An analytical chemist is titrating 116.9 mL of a 0.7700M solution of aniline (C6H5NH2) with a 0.5300M solution of HNO3. The pK of aniline is 9.37.
Calculate the pH of the base solution after the chemist has added 184.2 mL of the HNO 3 solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added.
Round your answer to 2 decimal places.
pH = ☐
☑
5
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