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Chapter 125 -- Part 3: Practice Test 3
The first thing to do is write the formulas for the two compounds.
Predict the products of the reaction.
When writing a formula, the positive and negative ion must be balanced.
The positive ion of the two compounds are exchanged in this type of reaction.
The products are PbI2 and NaNO3:Pb(NO3)2(aq) + 2NaI(aq) The number of moles of NaI added in the titration is shown.
moles of solute per liter is the definition of molarity.
Use the ratio from the equation to find the number of moles of lead.
Divide the number of moles by the volume of the sample to arrive at a figure.
The percent yield of a reaction is calculated by the formulaactual yield.
An increase in the theoretical yield of PbI2 would have been achieved.
The percent yield is the actual yield divided by the theoretical yield.
The mass of the sample and the mass of the compound are known.
We can use the mass of bromine:3.856 g total - 0.8568 g C - 0.1428 g H to calculate the number of moles of each element.
The mass of the empirical formula is not likely to be the actual formula since there is an odd number of total valence electrons available.
The mass is expected to be multiple of108.
The mass spectrum shows large peaks that are not a multiple of the number.
There are 3 small peaks near 216.
The compound has two isomers, 1,1-dichloroethane and 1,2-dichloroethane.
The hydrogen bonding is not present in the Lewis structures.
The 1,1-dichloroethane is more polar than the 1,2-dichloroethane, and the strong London dispersion forces strengthen the attraction between many 1,2-dichloroethane molecules.
The boiling point of 1,2-dichloroethane is higher.
The opposite poles of the molecule need to be aligned.
There are no extensive networks of attraction between Molecules.
The result is a lower boiling point.
The amount of Fe(III) present was not calculated by the first titration.
The second titration had all of the iron in the oxidation state after the reduction of Fe(III).
An oxidation is the loss of electrons to another substance.
The oxidation half-reaction involves the loss of one electron.
When writing an oxidation half-reaction, the electrons are placed on the product side.
Write two Cr3+ as products since there are two chromium ion in the reactant.
This reaction is not yet balanced, so begin by writing the species involved.
In order to balance the oxygen atoms involved, use water and hydrogen ion.
Adding 1 H2O for each O will balance the oxygen.
14 H+ on the reactant side will balance the charge.
Adding six electrons to the reactant side will give both sides a total charge of +6:6e- + Cr2O72-(aq) + 14H+(aq) - 2Cr3+(aq) + 7H2O.
The number of electrons gained and lost must be the same.
Adding the oxidation half-reaction to the reduction half-reaction will equal the 6 electrons in the reduction half-reaction.
Make sure that the charges on both sides are equal.
The mole ratio from the equation and the atomic mass of iron are the first things to be found.
The amount of Fe in the sample that was not measured can be found in the second titration.
The solution was prepared by dissolving the sample.
The mass of the sample was measured in half of the amount.
The Ksp can be written as: Co(OH)3(s) + Co3+(aq) + 3OH-(aq) Le Chatelier's principle suggests that the solubility of the solid in deionized water should decrease.
The presence of a common ion will shift the reaction toward the reactant, which will decrease the solubility of the salt.
The amount of hydroxide present in the solution is greater than 1.0 x 10-7 M. The water will form when the acid reacts with the hydroxide.
Stress to a system will shift in the direction to relieve it.
The removal of hydroxide ion by reaction with an acid will cause an increase in the molar solubility of the salt.
It is well known that converting a natural logarithm to a base 10 logarithm is done by dividing the ln by 2.303.
Since the antilog is 10 raised to the value, we can easily get the answer.
Since some reactions can be spontaneously based on the laws of nature, we make a distinction between them and nonspontaneous reactions.
A mixture of H2 and O2 without a spark is a good example.
The effective collision should have the chlorine atoms coming into contact with each other.
The first reaction is the slow step, so the Cl is not included in the overall reaction.
When carbonates react with acids, they form carbonic acid, which becomes carbon dioxide gas and liquid water.
The mole ratio from the net ionic equation is used to find the moles of calcium carbonate.
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