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Chapter 124 -- Part 2: Practice Test 2
A strong acid is completely ionized and the concentration is equal to the H+.
The only way to have a significant concentration of both the conjugate acid and conjugate base is through this method.
A polyprotic acid is indicative of an end point lower than 7.
The error is 1000 percent if 25degC is not converted to 298 K. The distribution curve shows that the highest point is at point D. The law of conserve energy is disobeyed by the other responses.
The opening valve 2 will increase the volume by 4 times, so the pressure will be 1/6 of the starting pressure.
The amount of particles is doubled when the second two valves are opened.
The element is 90 percent of an isotope with a mass of 20 and 10 percent.
The average mass of Ne is 20.2.
The weaker the acid is, the stronger it is.
The weak acid HF will form when NaF is dissolved in water.
The solution will be basic according to the following equation.
Choosing an indicator with the same pKa as the end point pH assures the sharpest color change and closest agreement of the end point (experimental) with the equivalence point.
Most of the indicator color change is outside the curve.
The titration of a weak acid has a slight jump at the beginning of the titration that strong acids do not.
Solid lines and hydrogen bonds are shown as dashed lines when aligned with the correct atoms.
The way to increase the reaction rate is to increase the ability of the reactants to meet.
It is easy to burn in a flame if the copper is grinding to a dust.
The equilibrium constant is only affected by temperature changes.
The square root of the Kb x concentration will be the OH-.
The square root is approximately 3 x 10-5.
The answer is between 10-5 and 10-4.
The pH will be between 9 and 10.
The Ka will be Kw/Kb.
The pKa will be -log.
The amount of material in each flask can be compared with the molar mass.
The lowest density is found in Flask A.
The higher the temperature, the less mass and density it has.
The flask with the highest temperature has the gases with the highest average energy.
The strongest intermolecular attractive forces are caused by the four bromine atoms with very polarizable electron clouds.
The London forces from the bromine atoms are the most important.
The dipole moments are small, and no hydrogen bonds are present.
Since the Lewis structure shows three bonding domains and four bonds, three sigma and one pi, the bond order is 4/3, not 5/3.
The sulfite ion has an unsymmetrical triangular pyramid structure.
This is the correct net ionic equation.
The lead and chloride ion charges are correct.
The spectator ion have been canceled.
The Ksp calculation shows that no precipitate should form.
This is a type of calculation.
The Ksp equation gives:Ksp is 1.7 x 10-5 andDividing by 0.17 gives4x2 and 1 x 10-4 The NO2 must increase when the reaction occurs.
All responses are correct.
If the reaction goes to completion, the correct pressure is 1520 torr because the reaction shows that there will be more moles of product than reactant.
The reaction doesn't go to completion since the pressure isn't reached.
The pressure of N2O4 will change.
The pressure in the vessel is 960 torr and the pressure of N2O4 is 760 torr.
We need to convert torr to atm by dividing by 760 torr/atm before entering values into the equation.
Only (B) has the correct form to fit the equation.
When more NO2 is injected, we expect to see a rise in pressure.
There will be a slower decrease due to the slow conversion of NO2 to N2O4.
It is not an important factor in solution formation.
The difference in boiling points is explained by the polarizability of the electron cloud.
Most of the edoic acid molecule are part of the dimers.
When substances are in the liquid state, heat must be added.
This is the definition of an endothermic process.
The choices are expected to be exothermic.
The affinity for fluorine is exothermic.
A formation reaction has elements as reactants and one formula unit of the product.
The only way to meet the definition is with a response.
The heat of reaction can be calculated if the temperature, standard cell potential, and equilibrium constant are known.
The sum of the two equations is what the desired equation is.
The sum of the two heats of reaction is -280.0 kJ.
If 0.200 mol of N2 reacts, we need 0.200 mol of O2.
N2 is the limiting reactant.
The equation q was used to calculate the heat evolved.
The minus sign shows heat is involved.
The first cell's reaction is Zn + Pb, the second cell's reaction isMg + Pb, and the third cell's reaction is Pb.
The cell potential is negative.
This was shown in the answer 42:Mg + Zn2+ - Mg2+ + Zn(E.K.
The voltage will decrease if Edegcell is positive and the product concentration is larger than the reactant concentration.
The voltage will increase if the product concentration is less than the reactant concentration.
This will happen in both cells.
Magnesium is the easiest oxidizer because both zinc and lead oxidize it.
Zinc is the second easiest oxidizer.
The periodic trends show that nitrogen is expected to be less than the preceding carbon.
The ionization energy for nitrogen should be larger than that of carbon because it increases from left to right across a period.
Perchloric acid is 100 percent ionized and is a strong acid.
Those with longer bond lengths are stronger acids.
The first three substances oxidize and cause other substances to be reduced, the opposite of what we want.
It is possible to replace chlorine with KMnO4.
Since we can see that ln A vs time produces a straight-line graph, the reaction is first order.
Half of the previous concentration remains at each hour, which is a characteristic of a first-order reaction.
The time needed for half of the reactant to react is defined as the half-life.
Since the value of t1/2 is known, k can be calculated.
It is possible to write [H+] as 10-3.89, which is also equal to [A]Ka.
The result will be -3.89 + (3.89) - (-2).
Take the negative of the logarithm to get the pKa.
The result was 5.78.
This response is correct and obeys the law of mass.
The atoms we have are 5O2 and 7S.
Stronger acids meanaker bonds to hydrogen.
A buffer must have a conjugate acid-base pair.
The strong acid neutralizing part of a weak base solution is the only solution that can produce this combination.
The other statements are misinterpreting the data.
The definition of a Bronsted-Lowry conjugate acid-base pair requires one H+.
This is the only reaction that shows two different phases.
The free energy can never be negative using these numbers.
NO3 is not a catalyst.
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