Chemistry High School

when one of the ions of the compound is already present in solution, its concentration at equilibrium will be higher, therefore making ksp larger.
True False

Answers

Answer 1

The statement "when one of the ions of the compound is already present in solution, its concentration at equilibrium will be higher, therefore making Ksp larger" is false, because Ksp remains constant if the concentration at equilibrium Will be higher.

The presence of one of the ions in the solution does not make the Ksp larger. The Ksp (solubility product constant) is a fixed value for a particular compound at a specific temperature, and it does not change based on the concentration of the ions in the solution. The ion concentrations may affect the position of the equilibrium, but the Ksp value remains constant.

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Related Questions

there are 8 isomeric alcohols with the formula C5H12O. draw the structure of this isomer: 3-methyl-2-butanol.

Answers

The formula C5H12O indicates that there are 5 carbon atoms, 12 hydrogen atoms, and 1 oxygen atom in the molecule.

To draw the structure of 3-methyl-2-butanol, we need to know that the name tells us there is a methyl (CH3) group on the third carbon atom, and that the molecule is a type of alcohol (ending in -ol) with a total of four carbon atoms in a chain, with a hydroxyl (-OH) group attached to the second carbon atom.

To draw the structure, we start by drawing a chain of four carbon atoms, with the second carbon atom having the -OH group attached to it. Then we add a methyl group (CH3) to the third carbon atom. Finally, we add enough hydrogen atoms to satisfy the valences of each atom, keeping in mind that each carbon atom needs four bonds and each hydrogen atom needs one bond. The resulting structure looks like this:

     CH3
      |
   H--C--OH
      |
   H--C--H
      |
   H--C--H
      |
      H

This is the structure for 3-methyl-2-butanol, which is one of the eight isomeric alcohols with the formula C5H12O.

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The following statements concern techniques used in the Titrationsexperiment. Select all the correct answers below.
A. If you rinse your buret with DI water, butdo not condition with the titrant solution, the effect is...
B. a decrease in the volume of the titrantrequired to reach the end point.
C. an increase in the volume of the titrantrequired to reach the end point.
D. an underestimation of the number ofmoles of analyte present.
E. an overestimation of the number of molesof analyte present.

Answers

The correct statements are:

A. If you rinse your buret with DI water, but do not condition with the titrant solution, the effect is...
C. an increase in the volume of the titrant required to reach the end point.
D. an underestimation of the number of moles of analyte present.

What is the effective way of conducting a Titration experiment?

When you rinse your buret with DI water but don't condition it with the titrant solution, the residual DI water in the buret will dilute the titrant solution. This dilution will cause an increase in the volume of the titrant required to reach the endpoint. Consequently, this will lead to an underestimation of the number of moles of analyte present in the solution.

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calculate the change in enthalpy associated with the combustion of 14.6 g of isooctane.

Answers

The change in enthalpy associated with the combustion of 14.6 g of isooctane is -672.7 kJ.


The first step in calculating the change in enthalpy associated with the combustion of isooctane is to write out the balanced chemical equation for the reaction:
C8H18 + 25/2 O2 -> 8 CO2 + 9 H2O
Next, we need to look up the standard enthalpy of formation (ΔHf°) values for each of the reactants and products in the equation. These values represent the change in enthalpy that occurs when one mole of the substance is formed from its constituent elements, under standard conditions (25°C and 1 atm pressure). Here are the relevant values:
ΔHf° (kJ/mol):
C8H18 = -258.8
O2 = 0
CO2 = -393.5
H2O = -285.8
Using these values, we can calculate the change in enthalpy (ΔH) for the combustion of 1 mole of isooctane:
ΔH = (8 x ΔHf°(CO2) + 9 x ΔHf°(H2O)) - (ΔHf°(C8H18) + 25/2 x ΔHf°(O2))
ΔH = (8 x -393.5 kJ/mol + 9 x -285.8 kJ/mol) - (-258.8 kJ/mol + 25/2 x 0 kJ/mol)
ΔH = -5515.7 kJ/mol + 258.8 kJ/mol
ΔH = -5256.9 kJ/mol
So, the change in enthalpy for the combustion of 1 mole of isooctane is -5256.9 kJ/mol. To find the change in enthalpy for the combustion of 14.6 g of isooctane, we need to convert the mass of isooctane to moles using its molar mass (114.23 g/mol):
n = 14.6 g / 114.23 g/mol
n = 0.128 mol
Now we can use the calculated ΔH value to find the change in enthalpy for the combustion of this amount of isooctane:
ΔH = -5256.9 kJ/mol x 0.128 mol
ΔH = -672.7 kJ

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