This web page has every one of the compelled homework because that the material covered in the second exam that the an initial semester of basic Chemistry. The textbook connected with this homework is CHEMISTRY The main Science by Brown, LeMay, et.al. The critical edition I compelled students come buy to be the 12th version (CHEMISTRY The central Science, 12th ed. By Brown, LeMay, Bursten, Murphy and Woodward), but any kind of edition the this message will carry out for this course.

Note: You space expected to go to the end of chapter problems in your textbook, find similar questions, and also work the end those difficulties as well. This is simply the required perform of troubles for quiz purposes. Friend should also study the Exercises in ~ the chapters. The exercises are cleared up examples of the concerns at the earlier of the chapter. The examine guide likewise has resolved examples.

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These room bare-bones questions. The textbook questions will have additional information that may be useful and that connect the troubles to genuine life applications, many of them in biology.

Aqueous reactions (Chapter Four)What ions will be existing when every of the following substances is inserted in water?BaCl2AnswerBa2+ and also Cl-AgClAnswerAgCl is considered insoluble, for this reason our answer would be the no ions are formed.ZnSO4AnswerZn2+ and also SO42-HNO3AnswerH+ and NO3-Na2CO3AnswerNa+ and also CO32-BaSO4AnswerBaSO4 is thought about insoluble, so us say the no ions space formed.NaOHAnswerNa+ and OH-What will be liquified in the solution when each of the adhering to weak acids are inserted in water?Hypochlorous acid, HClOAnswerH+, ClO- and also HClOPropionic acid, C2H5COOH or HC3H5O2AnswerH+, C2H5COO- and also C2H5COOHWrite the molecular, ionic and net ionic equations when the complying with substances are put together in one aqueous solution.CaBr2 and also K2CO3AnswerCaBr2(aq) + K2CO3(aq) → CaCO3(s) + 2KBr(aq)Ca2+(aq) + 2Br-(aq) + 2K+(aq) + CO32-(aq) → CaCO3(s) + 2K+(aq) + 2Br-(aq)Ca2+(aq) + CO32-(aq) → CaCO3(s)CdSO4 and also Ba(C2H3O2)2AnswerCdSO4(aq) + Ba(C2H3O2)2(aq) → BaSO4(s) + Cd(C2H3O2)2(aq)Cd2+(aq) + SO42-(aq) + Ba2+(aq) + 2C2H3O2-(aq) → BaSO4(s) + Cd2+(aq) + 2C2H3O2-(aq)Ba2+(aq) + SO42-(aq) → BaSO4(s)HNO3 and also PbCO3Answer2HNO3(aq) + PbCO3(s) → CO2(g) + H2O(l) + Pb(NO3)2(aq)2H+(aq) + 2NO3-(aq) + PbCO3(s) → CO2(g) + H2O(l) + Pb2+(aq) 2NO3-(aq)2H+(aq) + PbCO3(s) → CO2(g) + H2O(l) + Pb2+(aq)(NH4)3PO4 and CuCl2Answer2(NH4)3PO4(aq) + 3CuCl2(aq) → Cu3(PO4)2(s) + 6NH4Cl(aq)6NH4+(aq) + 2PO43-(aq) + 3Cu2+(aq) + 6Cl-(aq) → Cu3(PO4)2(s) + 6NH4+(aq) + 6Cl-(aq)3Cu2+(aq) + 2PO43-(aq) → Cu3(PO4)2(s)MgBr2 and also CaSO4AnswerMgBr2(aq) + CaSO4(aq) → CaBr2(aq) + MgSO4(aq)Mg2+(aq) + 2Br-(aq) + Ca2+(aq) + SO42-(aq) → Mg2+(aq) + 2Br-(aq) + Ca2+(aq) + SO42-(aq)No network Ionic Equation.A solution may contain any or every one of the following ions: Ba2+, Ni2+, or Pb2+. From the following information, i beg your pardon ions room present? display reasoning.When HCl is included to the systems a precipitate is formed. The precipitate is filtered from the solution. As soon as K2SO4 is added to the filtered systems no precipitate is formed. As soon as NaOH is then added to the filtered systems a precipitate is formed.AnswerA reaction table mirroring what precipitates are feasible would look choose the following:
Ba2+Ni2+Pb2+
HClBaCl2(aq)NiCl2(aq)PbCl2(s)
K2SO4BaSO4(s)NiSO4(aq)--
NaOH--Ni(OH)2(s)--

A precipitate through HCl method that Pb2+ must be present. No precipitate v K2SO4 method that Ba2+ can not it is in present. A precipitate v NaOH method that Ni2+ should be present.

Answer: Pb2+ and Ni2+ room in the solution.A solution may contain any or every one of the adhering to ions: Sr2+, Fe2+, or Ca2+. From the following information, which ions space present? present reasoning.When KOH is added to the equipment a precipitate is formed. The precipitate is filtered native the solution. Once Na2SO4 is added to the filtered solution a precipitate is again formed. This brand-new precipitate is then filtered native the solution. When (NH4)2CO3 is then added to the filtered equipment a precipitate is not formed.AnswerA reaction table reflecting what precipitates are possible would look favor the following:
Sr2+Fe2+Ca2+
KOHSr(OH)2(aq)Fe(OH)2(s)Ca(OH)2(aq)
Na2SO4SrSO4(s)--CaSO4(aq)
(NH4)2CO3----CaCO3(s)
A precipitate with KOH means that Fe2+ have to be present. A precipitate v Na2SO4 way that Sr2+ is present. No precipitate v NaOH method that Ca2+ is no present.

Answer: Sr2+ and Fe2+ are in the solution.Answer the following oxidation/reduction associated questions:What facet has the strongest attraction because that electrons within a bond?AnswerFlourine, FWhat is the attraction because that electrons within a link called?AnswerElectronegativityCan oxidation happen without reduction?AnswerNoWhat execute we know about the number of electrons lost compared to the number of electrons gained?AnswerThey must be equal.What is the tendency for electronegativity throughout and down the regular table?AnswerThe electronegativity increases going native left to right and also decreases going under the routine table.True or False: The assignment that oxidation numbers assumes that the most electronegative facet will get the electrons, or in various other words that the bonds will be favor ionic bonds. AnswerTrue.What is the oxidation number of the alkali metals?Answer1+What is the oxidation variety of flourine?Answer1-What is the oxidation variety of a single element ion?AnswerThe charge on the ion.What is the oxidation number of elements in their conventional state?AnswerZero.What is the oxidation number preeminence for polyatomic ions?AnswerAll of the oxidation numbers must include up come the fee on the ion.What is the oxidation number ascendancy for molecules that don"t have any charge?AnswerThe oxidation number must include up to zero.What is the normal oxidation number of oxygen?Answer2-, but there are some exceptions.What is the oxidation variety of each aspect in every of the following substances?HClOAnswerH: +1, O: -2, Cl: +1C2H5OHAnswerH: +1, O: -2, C: -2HSO4-AnswerH: +1, O: -2, S: +6PO43-AnswerO: -2, P: +5Cl2(g)AnswerCl: 0 (zero)COCl2AnswerCl: -1, O: -2, C: +4KMnO4AnswerK: +1, O: -2, Mn: +7Al(NO3)3AnswerO: -2, N: +5, Al: +3For each of the following reactions: (a) use the oxidation number method to balance the reaction (show work, i.e., oxidation numbers, electrons transferred, etc.). (b) recognize the facet that is reduced and also the aspect that is oxidized. (c) identify the oxidizing agent and the reduce agent.Cd(s) + Ag+(aq) → Cd2+(aq) + Ag(s)Answer

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Fe(NO3)2(aq) + Al(s) → Fe(s) + Al(NO3)3(aq)Answer
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Cr2+(aq) + MnO4-(aq) + H+(aq) → Cr3+(aq) + Mn2+(aq) + H2O(l)Answer
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P4(s) + HClO(aq) + H2O → H3PO4(aq) + HCl(aq)Answer
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C2H5OH(l) + O2(g) → H2O(l) + CO2(g)Answer
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Use the fifty percent reaction method to complete and balance the following reactions.SO2 + NO3- → SO42- + NO (acidic solution)Answer
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ClO3- + Br- → Cl2 + Br2 (acidic solution)Answer
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Cr2+ + MnO4- → Cr3+ + Mn2+ (acidic solution)Answer
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Sn(OH)42- + BrO- → SnO2 + Br- (basic solution)Answer
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S- + NO3- → S + NO (basic solution)Answer
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I- + OCl- → IO3- + Cl-Answer
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NO2- + Ga → NH4+ + GaO2-Answer
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Answer every of these molarity problems.What is the meaning of molarity?Answermolarity is characterized as mole of solute split by liters of solution.What is the difference in between 0.4 mol and 0.4 M?Answer0.4 mol is the number of moles when 0.4 M is the number of moles per liter of solution.What is the molarity as soon as 1.2 mole of NaCl is put in a systems where the total volume is 2.4 L?Answer
molarity = 1.2 mole NaCl
2.4 L soln
= 0.5 mole NaCl
1 L soln
= 0.5 MWhat is the molarity when 5.0 g the KMnO4 are placed in water to make a complete solution having actually a volume of 500 mL?Answer
( 5 g KMnO4 )( 1 mole KMnO4
158.04 g KMnO4
) = 0.0316 mole KMnO4

molarity = 0.0316 mole KMnO4
0.5 L soln
= 0.0633 mole KMnO4
1 L soln
= 0.0633 M KMnO4How numerous moles the KCl room in 250 mL the a 1.2 M KCl solution?Answer
( 0.25 L soln )( 1.2 mole KCl
1 L soln
) = 0.3 moles KClWhat volume of 2.2 M LiBr is needed to have 1.1 mole the LiBr?Answer
( 1.1 mole LiBr )( 1 L soln
2.2 mole LiBr
) = 0.5 together soln How countless grams of NaOH are required to do 500 mL of a 1.5 M NaOH solution?Answer
( 0.5 L soln )( 1.5 mole NaOH
1 L soln
)( 39.99 g NaOH
1 mole NaOH
) = 29.99 g NaOHWhat concentration need to a stock solution be in bespeak to use 10 mL of it come prepare 250 mL of a 0.2 M solution?Answer
( 0.250 L soln )( 0.2 mole solute
1 L soln
) = 0.05 mole solute

0.05 mole solute
0.010 L soln
= 5 MHow lot of a 1.5 M stock solution must you use to prepare 100 mL of a 0.5 M solution?Answer
( 0.1 L soln )( 0.5 mole solute
1 L soln
)( 1 L stock soln
1.5 mole solute
) = 0.033 L share soln

0.033 together or 33 mL that the share soln room neededA condition called hyponatremia occurs when the sodium concentration in the blood falls listed below 0.130 M. A typical adult has 4.7 l of blood. (a) How numerous grams the sodium room in the blood of a usual adult? (b) how much sodium would a person need to gain into your blood if the sodium concentration was 0.12 M?Answer
( 4.7 L blood )( 0.130 mole Na
1 L blood
)( 22.99 g Na
1 mole Na
) = 14.05 g Na (a)

( 4.7 L blood )( 0.120 mole Na
1 L blood
)( 22.99 g Na
1 mole Na
) = 12.97 g Na

Amount essential = 14.05 g - 12.97 g = 1.08 g Na needed (b)Many places have actually laws that say a person is legally drunk as soon as they have actually a concentration of much more than 0.013 M in ethanol, C2H5OH. A common adult has 4.7 l of blood. (a) How countless grams of ethanol are in the blood the a typical adult once legally drunk? (b) A have the right to of beer frequently has about 20 g of ethanol. How plenty of cans the beer will administer the legitimate limit?Answer
( 4.7 L blood )( 0.013 mole C2H5OH
1 L blood
)( 62 g C2H5OH
1 mole C2H5OH
) = 3.79 g C2H5OH (a)

( 3.79 g C2H5OH )( 1 can
20 g C2H5OH
) = 0.19 can just one 5th of a can! (b)What massive of Na3PO4 is needed to fully precipitate the Ni2+ ion that space in 15 mL of a solution that is 0.8 M NiCl2?Answer2Na3PO4(aq) + 3NiCl2(aq) → Ni3(PO4)2(s) + 6NaCl(aq)

( 0.015 L NiCl2 solution )( 0.8 mole NiCl2
1 L NiCl2 solution
) = 0.012 mole NiCl2

( 0.012 mole NiCl2 )( 2 mole Na3PO4
3 mole NiCl2
)( 164 g Na3PO4
1 mole Na3PO4
) = 1.3 g Na3PO4How many mL the 0.5 M NaOH is needed to fully neutralize 100 mL that 0.2 M HCl?AnswerNaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

( 0.1 L HCl soln )( 0.2 mole HCl
1 L HCl soln
) = 0.02 mole HCl

( 0.02 mole HCl )( 1 mole NaOH
1 mole HCl
)( 1 L NaOH soln
0.5 mole NaOH
) = 0.04 l or 40 mL NaOH soln15 mL that 0.1 M NaOH is required to fully neutralize 20 mL of an H2SO4 solution. What was the concentration that the original H2SO4 solution?Answer2NaOH(aq) + H2SO4(aq) → 2H2O(l) + Na2SO4(aq)

( 0.015 L NaOH soln )( 0.1 mole NaOH
1 L NaOH soln
) = 0.0015 mole NaOH

( 0.0015 mole NaOH )( 1 mole H2SO4
2 mole NaOH
) = 0.00075 mole H2SO4

= 0.00075 mole H2SO4
0.02 L soln
= = 0.0375 M H2SO4Assume a silver acetate solution that is 1.4 M AgC2H3O2 when doing the complying with problems.Write the balanced chemical equation because that the reaction of silver acetate v sodium chloride in an aqueous solution.AnswerAgC2H3O2(aq) + NaCl(aq) → AgCl(s) + NaC2H3O2(aq)How plenty of grams that NaCl are essential to fully react with 200 mL of the AgC2H3O2 solution?Answer
( 0.2 L AgC2H3O2 solution )( 1.4 mole AgC2H3O2
1 L AgC2H3O2 solution
) = 0.28 mole AgC2H3O2

( 0.28 mole AgC2H3O2 )( 1 mole NaCl
1 mole AgC2H3O2
)( 58.5 g NaCl
1 mole NaCl
) = 16.4 g NaClHow lot silver chloride is created when 100 mL that the AgC2H3O2 systems is linked with 200 mL the 0.6 M NaCl?AnswerDetermine the moles of every reactant. Us have:

( 0.1 L AgC2H3O2 soln )( 1.4 mole AgC2H3O2
1 L AgC2H3O2 soln
) = 0.14 mole AgC2H3O2

( 0.2 L NaCl soln )( 0.6 mole NaCl
1 L NaCl soln
) = 0.12 mole NaCl

Determine limitiing reactant by selecting one reactant and determining the quantity of the various other one the is required to fully react. I have chosen NaCl below.

( 0.12 mole NaCl )( 1 mole AgC2H3O2
1 mole NaCl
) = 0.12 mole AgC2H3O2 needed

We need 0.12 mole AgC2H3O2, yet have 0.14 mole. We have actually extra AgC2H3O2 and so NaCl is the limiting reactant. Usage the lot of limiting reactant that we have to determine the lot of product produced.

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( 0.12 mole NaCl )( 1 mole AgCl
1 mole NaCl
)( 143.4 g AgCl
1 mole AgCl
) = 17.2 g AgCl What are the concentration of each of the ion in the final solution as soon as 100 mL of the AgC3H3O2 systems is an unified with 200 mL the 0.6 M NaCl? Use info calculated in the critical problem and assume the volumes space additive (the complete volume is the amount of the volumes that are added together).AnswerFrom the last trouble we watch that the NaCl to be the limiting reactant. That means that all of the Cl will certainly be in AgCl(s) and the concentration the the Cl- is zero: = 0.

There to be extra Ag+. The amount offered was 0.12 moles, but we had 0.14 moles. That means that there to be 0.02 mole extra that would continue to be in the systems after the reaction was complete. Those 0.02 moles would certainly be in a total of 300 mL, so
= 0.02 mole Ag+
0.3 L soln
= 0.067 mole Ag+
1 L soln
= 0.067 M Ag+

Na+ doesn"t do a precipitate, so all of the original amount would be in the last solution and
= 0.12 mole Na+
0.3 L soln
= 0.4 mole Na+
1 L soln
= 0.4 M Na+

C2H3O2- doesn"t make a precipitate, so every one of the original amount would certainly be in the final solution and
= 0.14 mole C2H3O2-
0.3 L soln
= 0.467 mole C2H3O2-
1 L soln
= 0.467 M C2H3O2-A 100 mL systems of 1.2 M KOH is merged with 150 mL the 1.0 M ZnCl2.How countless grams of Zn(OH)2 can be formed?HintWhat is the well balanced chemical equation?How numerous moles of every reactant perform you begin with? (Use molarity.)What is the limiting reactant? (Use mole ratios from balanced eqn.)How many moles of product are formed? (Use mole ratios from well balanced eqn.)How plenty of grams the product space formed? (Use routine table.)How numerous moles the each original ion is left over?What is the full volume?What is the last molarity of each ion? (Moles left over divided by total volume.)Answer2KOH(aq) + ZnCl2(aq) → 2KCl(aq) + Zn(OH)2(s)

Determine the mole of every reactant. We have:

( 0.1 L KOH soln )( 1.2 mole KOH
1 L KOH soln
) = 0.12 mole KOH

( 0.15 L ZnCl2 soln )( 1.0 mole ZnCl2
1 L ZnCl2 soln
) = 0.15 mole ZnCl2

Determine limitiing reactant by picking one reactant and also determining the amount of the various other one that is needed to totally react. Making use of KOH:

( 0.12 mole KOH )( 1 mole ZnCl2
2 mole KOH
) = 0.06 mole ZnCl2 needed

We require 0.06 mole ZnCl2, however have 0.15 mole. We have actually extra ZnCl2 and so KOH is the limiting reactant. Use the lot of limiting reactant the we have to determine the quantity of product produced.

( 0.12 mole KOH )( 1 mole Zn(OH)2
2 mole KOH
)( 99.4 g Zn(OH)2
1 mole Zn(OH)2
) = 5.96 g Zn(OH)2 What are the concentrations of each of the ion in the final solution? assume the volumes space additive (the full volume is the amount of the quantities that are added together).AnswerFrom the last difficulty we view that the KOH was the limiting reactant. That means that all of the OH- will be in Zn(OH)2(s) and the concentration the the OH- is zero: = 0.

There to be extra Zn2+. The amount used was 0.06 mole ZnCl2, but we had actually 0.15 moles. That method that there to be 0.15 - 0.06 = 0.09 moles extra Zn2+ the would continue to be in the equipment after the reaction to be complete. Those 0.09 moles would certainly be in a total of 250 mL, so
= 0.09 mole Zn2+
0.25 L soln
= 0.36 mole Zn2+
1 L soln
= 0.36 M Zn2+

K+ doesn"t do a precipitate, so every one of the initial amount, 0.12 mole KOH, would certainly be in the last solution and
= 0.12 mole K+
0.25 L soln
= 0.48 mole K+
1 L soln
= 0.48 M K+

Cl- doesn"t do a precipitate, so every one of the initial amount, (0.15 mole ZnCl2)(2 mole Cl-/mole ZnCl2) = 0.3 mole Cl-, would be in the final solution and
= 0.3 mole Cl-
0.25 L soln
= 1.2 mole Cl-
1 L soln
= 1.2 M Cl-