This page has all of the compelled homework for the material covered in the second exam of the an initial semester of general Chemistry. The textbook linked with this homework is CHEMISTRY The central Science by Brown, LeMay, The last edition I forced students come buy to be the 12th edition (CHEMISTRY The central Science, 12th ed. By Brown, LeMay, Bursten, Murphy and Woodward), but any kind of edition of this text will do for this course.

Note: You room expected to go to the end of chapter troubles in your textbook, find comparable questions, and work the end those problems as well. This is simply the required perform of difficulties for quiz purposes. Girlfriend should additionally study the Exercises within the chapters. The exercises are worked out examples that the inquiries at the back of the chapter. The examine guide additionally has resolved examples.

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These room bare-bones questions. The textbook questions will have added information that may be useful and also that associate the difficulties to genuine life applications, countless of them in biology.

Aqueous reactions (Chapter Four)What ions will be current when every of the adhering to substances is put in water?BaCl2AnswerBa2+ and also Cl-AgClAnswerAgCl is taken into consideration insoluble, so our answer would be the no ions room formed.ZnSO4AnswerZn2+ and also SO42-HNO3AnswerH+ and also NO3-Na2CO3AnswerNa+ and also CO32-BaSO4AnswerBaSO4 is taken into consideration insoluble, so we say that no ions space formed.NaOHAnswerNa+ and also OH-What will certainly be liquified in the solution when each the the complying with weak mountain are put in water?Hypochlorous acid, HClOAnswerH+, ClO- and also HClOPropionic acid, C2H5COOH or HC3H5O2AnswerH+, C2H5COO- and also C2H5COOHWrite the molecular, ionic and also net ionic equations as soon as the adhering to substances are put together in one aqueous solution.CaBr2 and 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 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 adhering to ions: Ba2+, Ni2+, or Pb2+. From the complying with information, i m sorry ions are present? show reasoning.When HCl is added to the equipment a precipitate is formed. The precipitate is filtered native the solution. As soon as K2SO4 is included to the filtered equipment no precipitate is formed. When NaOH is then added to the filtered solution a precipitate is formed.AnswerA reaction table mirroring what precipitates are feasible would look choose the following:

A precipitate through HCl means that Pb2+ need to be present. No precipitate with K2SO4 method that Ba2+ have the right to not it is in present. A precipitate with NaOH means that Ni2+ must be present.

Answer: Pb2+ and Ni2+ space in the solution.A solution might contain any type of or every one of the following ions: Sr2+, Fe2+, or Ca2+. Native the complying with information, which ions space present? present reasoning.When KOH is included to the solution a precipitate is formed. The precipitate is filtered from the solution. When Na2SO4 is included to the filtered equipment a precipitate is again formed. This brand-new precipitate is climate filtered from the solution. Once (NH4)2CO3 is then included to the filtered solution a precipitate is not formed.AnswerA reaction table reflecting what precipitates are feasible would look like the following:
A precipitate with KOH means that Fe2+ should be present. A precipitate through Na2SO4 way that Sr2+ is present. No precipitate v NaOH method that Ca2+ is no present.

Answer: Sr2+ and Fe2+ space in the solution.Answer the adhering to oxidation/reduction connected questions:What element has the the strongest attraction for electrons within a bond?AnswerFlourine, FWhat is the attraction because that electrons within a link called?AnswerElectronegativityCan oxidation happen without reduction?AnswerNoWhat do we know about the number of electrons lost contrasted to the variety of electrons gained?AnswerThey have to be equal.What is the tendency for electronegativity throughout and down the routine table?AnswerThe electronegativity boosts going from left to right and decreases going down the routine table.True or False: The assignment the oxidation number assumes the the many electronegative facet will get the electrons, or in other words that the bonds will be favor ionic bonds. AnswerTrue.What is the oxidation variety of the alkali metals?Answer1+What is the oxidation variety of flourine?Answer1-What is the oxidation number of a solitary element ion?AnswerThe charge on the ion.What is the oxidation number of elements in their traditional state?AnswerZero.What is the oxidation number dominance for polyatomic ions?AnswerAll the the oxidation number must add 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 element in each of the adhering to 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 every of the complying with reactions: (a) usage the oxidation number technique to balance the reaction (show work, i.e., oxidation numbers, electron transferred, etc.). (b) determine the facet that is reduced and the facet that is oxidized. (c) recognize the oxidizing agent and also the to reduce agent.Cd(s) + Ag+(aq) → Cd2+(aq) + Ag(s)Answer

Fe(NO3)2(aq) + Al(s) → Fe(s) + Al(NO3)3(aq)Answer
Cr2+(aq) + MnO4-(aq) + H+(aq) → Cr3+(aq) + Mn2+(aq) + H2O(l)Answer
P4(s) + HClO(aq) + H2O → H3PO4(aq) + HCl(aq)Answer
C2H5OH(l) + O2(g) → H2O(l) + CO2(g)Answer
Use the half reaction technique to complete and also balance the following reactions.SO2 + NO3- → SO42- + NO (acidic solution)Answer
ClO3- + Br- → Cl2 + Br2 (acidic solution)Answer
Cr2+ + MnO4- → Cr3+ + Mn2+ (acidic solution)Answer
Sn(OH)42- + BrO- → SnO2 + Br- (basic solution)Answer
S- + NO3- → S + NO (basic solution)Answer
I- + OCl- → IO3- + Cl-Answer
NO2- + Ga → NH4+ + GaO2-Answer
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 between 0.4 mol and 0.4 M?Answer0.4 mol is the variety of moles when 0.4 M is the variety of moles every liter the solution.What is the molarity as soon as 1.2 moles of NaCl is put in a equipment where the full 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 that KMnO4 are put in water to do a total solution having actually a volume that 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 countless moles the KCl room in 250 mL that a 1.2 M KCl solution?Answer
( 0.25 L soln )( 1.2 mole KCl
1 L soln
) = 0.3 mole KClWhat volume that 2.2 M LiBr is necessary to have 1.1 mole of LiBr?Answer
( 1.1 mole LiBr )( 1 L soln
2.2 mole LiBr
) = 0.5 l soln How plenty of grams of NaOH are essential to do 500 mL the 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 equipment be in stimulate 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 much of a 1.5 M stock solution need to you usage to prepare 100 mL the a 0.5 M solution?Answer
( 0.1 L soln )( 0.5 mole solute
1 L soln
)( 1 L share soln
1.5 mole solute
) = 0.033 L share soln

0.033 l or 33 mL that the stock soln room neededA condition called hyponatremia occurs as soon as the salt concentration in the blood falls listed below 0.130 M. A typical adult has 4.7 l of blood. (a) How many grams of sodium room in the blood of a common adult? (b) exactly 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 laws the say a person is legitimate drunk when they have a concentration of an ext than 0.013 M in ethanol, C2H5OH. A usual adult has 4.7 together of blood. (a) How countless grams the ethanol room in the blood that a typical adult as soon as legally drunk? (b) A deserve to of beer generally has around 20 g the ethanol. How countless cans of beer will provide the legit 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 fifth of a can! (b)What mass of Na3PO4 is needed to fully precipitate the Ni2+ ion that room in 15 mL the a systems 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 countless mL the 0.5 M NaOH is necessary to fully neutralize 100 mL the 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 the 0.1 M NaOH is needed to totally neutralize 20 mL of one H2SO4 solution. What to be the concentration that the initial 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 systems that is 1.4 M AgC2H3O2 when doing the adhering to problems.Write the balanced chemical equation because that the reaction of silver acetate through sodium chloride in an aqueous solution.AnswerAgC2H3O2(aq) + NaCl(aq) → AgCl(s) + NaC2H3O2(aq)How numerous grams that NaCl are necessary to fully react with 200 mL that 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 produced when 100 mL of the AgC2H3O2 systems is combined with 200 mL that 0.6 M NaCl?AnswerDetermine the mole of each 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 choosing one reactant and also determining the quantity of the various other one the is necessary to totally react. I have actually chosen NaCl below.

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

We require 0.12 mole AgC2H3O2, yet have 0.14 mole. We have actually extra AgC2H3O2 and also so NaCl is the limiting reactant. Usage the quantity of limiting reactant the 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 space the concentration of each of the ions in the final solution as soon as 100 mL the the AgC3H3O2 systems is an unified with 200 mL that 0.6 M NaCl? Use details calculated in the last problem and assume the volumes room additive (the full volume is the amount of the quantities that are included together).AnswerFrom the last difficulty we check out that the NaCl was the limiting reactant. That way that every one of the Cl will be in AgCl(s) and also the concentration that the Cl- is zero: = 0.

There to be extra Ag+. The amount provided was 0.12 moles, however we had actually 0.14 moles. That method that there to be 0.02 mole extra the would stay in the systems after the reaction was complete. Those 0.02 moles would be in a full 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 final 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 initial 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 combined with 150 mL that 1.0 M ZnCl2.How countless grams that Zn(OH)2 can be formed?HintWhat is the well balanced chemical equation?How numerous moles of each reactant carry out you start with? (Use molarity.)What is the limiting reactant? (Use mole ratios from balanced eqn.)How many moles of product room formed? (Use mole ratios from balanced eqn.)How plenty of grams that product room formed? (Use periodic table.)How many moles that each original ion is left over?What is the total volume?What is the final molarity of each ion? (Moles left over split by total volume.)Answer2KOH(aq) + ZnCl2(aq) → 2KCl(aq) + Zn(OH)2(s)

Determine the moles of each reactant. Us 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 selecting one reactant and also determining the amount of the various other one that is needed to completely react. Making use of KOH:

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

We need 0.06 mole ZnCl2, however have 0.15 mole. We have actually extra ZnCl2 and so KOH is the limiting reactant. Usage the amount of limiting reactant that we need 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 room the concentration of each of the ions in the final solution? assume the volumes are additive (the total volume is the sum of the volumes that are included together).AnswerFrom the last problem we check out that the KOH to be the limiting reactant. That means that every one of the OH- will certainly be in Zn(OH)2(s) and also the concentration of the OH- is zero: = 0.

There to be extra Zn2+. The amount provided was 0.06 mole ZnCl2, but we had actually 0.15 moles. That way that there to be 0.15 - 0.06 = 0.09 mole extra Zn2+ that would remain in the solution after the reaction was complete. Those 0.09 moles would certainly be in a complete 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 be in the final 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 original amount, (0.15 mole ZnCl2)(2 mole Cl-/mole ZnCl2) = 0.3 mole Cl-, would certainly 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-