PCAT - General Chemistry Problems [IV-VI]

If 92 grams of glycerol is mixed with 90 grams of water, what will be the mole fractions of the two components? Note: (molecular mass of water = 18 g/mol & glycerol = 92 g/mol)

--> 90 g H2O = [90 g/1] ∙ [1 mol/18 g] = 5 mol. --> 92 g glycerol = [92 g/1] ∙ [1 mol/92 g] = 1 mol. --> Total mol = [1 + 5] = 6 mol. --> Xwater = [5 mol/6 mol] = 0.883. --> Xglycerol = [1 mol/6 mol] = 0.167. To double check, make sure you have a solid 1 for total: --> [0.167 + 0.833] = 1.000.

Aspirin is prepared by reacting salicylic acid & acetic anhydride as follows: C7H6O3 + C4H6O3 --> C9H8O4 + C2H4O2. How many moles of salicylic acid should be used to prepare six 5-grain aspirin tablets? Note: (1 gram = 15.5 grains).

1 mol salicylic acid will yield 1 mol aspirin. First, determine how many grams 5 grains are using the information from the question. [1 gram / 15.5 grains ] = [ x grams / 5 grains] x = 0.323 grams ∙ 6 tablets = 2 grams. Now we can solve the problem using basic stoichiometry. [2 grams aspirin/1]∙[1 mol aspirin/180 g aspirin]∙[1 mol S.A/1 mol aspirin] = 0.011 mol salicylic acid.

Given that the molecular mass of ethyl alcohol, CH3CH2OH is 46 g/mol, & that of water is 18 g/mol, how many grams of ethyl alcohol must be mixed with 100 mL of water for the mole fraction (X) of ethyl alcohol to be 0.2?

64.4 g. The # of moles of H2O is found by estimating the density of water to be 1 g/mL. --> mol H2O = [(1,00 mL H2O)(1 g/mL)/(18.0 g/mol)] = 5.6 mol. If mole fraction of ethyl alcohol is to be 0.2, mole fraction of H2O must be 0.8. If n = the total # of moles: --> 5.6 mol = 0.8n. --> n = 7 mol. Then: --> Mole ethyl alcohol = (0.2)(7 mol) = 1.4 mol. &: --> Grams ethyl alcohol = (1.4 mol ethyl alcohol)(46 g/mol) = 64.4 g ethyl alcohol.

The reaction below is best classified as which type of reaction? CH3CO2Na + HClO4 --> CH3CO2H + NaClO4. A. Double-Displacement Reaction. B. Combination Reaction. C. Decomposition Reaction. D. Single-Displacement & Decomposition Reaction.

A. Because the only change is that the Na from CH3CO2Na exchanges with the H from HClO4, this is a double displacement rxn. Alternately, this rxn could be classified as a neutralization, since it is a rxn between an acid & base.

The activation energy for a reaction in the forward direction is 78 kJ. The activation energy for the same reaction in reverse is 300 kJ. If the energy of the products is 25 kJ, then: A. What is the energy of the reactant? B. Is the forward reaction endothermic or exothermic? C. Is the reverse reaction endothermic or exothermic? D. What is the enthalpy change for the forward reaction?

A. Draw a diagram [see attached]. --> ΔH = Ea[→] - Ea[←]. --> ΔH = 78 kJ - 300 kJ = -222 kJ. & ΔH= E[Products] - E[Reactants], so: --> -222 kJ = 25 kJ - X --> X = 247 kJ = E[Reactants]. B. → is exothermic because ΔH is (-). C. ← is endothermic because ΔH is (+). D. The enthalpy change (ΔH) is -222 kJ, (from A).

How much NaOH must be added to 200 mL of water to make a 1.0 M NaOH solution? A. 8.0 g. B. 16 g. C. 40 g. D. 80 g.

A. First, molar mass of NaOH = 40 g/mol via the periodic table. Thus, 40 g of NaOH = 1 mol, & 40 g of NaOH dissolved in 1 L of water = 1 M solution of NaOH. We can use the following ratio to solve: --> [x/200 mL] = [40 g/1 L] ∙ [1 L/1,000 mL]. --> [x/200 mL] = [40 g/1,000 mL] --> x = [(40 g ∙ 200 mL)/(1,000 mL)] = 8 g.

Consider the following chemical reaction & experimental data: A (aq) --> B (aq) + C (g). Trial I: [A]: 0.10, 0.20, 0.30, 0.40. Rate: 0.6, 0.6, 0.6, 0.6. Trial II: [A]: 0.10, 0.20, 0.30, 0.40. Rate: 0.9, 0.9, 0.9, 0.9. A. What is the rate expression for Trial I? B. What is the rate constant for Trial I? C. What is most likely the reason for the increased rate in Trial II?

A. Rate = k[A]0= k. This rxn has only 1 reactant. It is evident from the data that the rate of the rxn is not affected by reactant concentration. This is, then, a 0-order rxn, & the rate = k. B. k= 0.6. C. The most likely reason for the increased rate in trial 2 is change in T. We know that in a 0-order rxn, changing the concentration of the reactant will not cause a change in rate.

Name the following ionic compounds: A. NaClO4. B. NaClO. C. NaNO3. D. KNO2. E. Li2SO4. F. MgSO3.

A. Sodium perchlorate. B. Sodium hypochlorite. C. Sodium nitrate. D. Potassium nitrate. E. Lithium sulfate. F. Magnesium sulfite.

Consider the following experimental data: SO3 + H2O --> H2SO4. Trial I: [SO3]: 0.1. [H2O]: 0.01. Rate: 0.013. Trial II: [SO3]: 0.2. [H2O]: 0.01. Rate: 0.052. Trial III: [SO3]: X. [H2O]: 0.02. Rate: 0.234. Trial IV: [SO3]: 0.1. [H2O]: 0.03. Rate: 0.039. A. What is the value of X? B. What is the order of the reaction? C. What is the rate constant? D. What would be the rate if [SO3] in Trial IV were raised to 0.2?

A. X = 0.3. To calculate X, first write the rate expression from the table given [see attached]: rate = k[SO3]^2[H2O]^1. The order with respect to SO3 is 2, since rate quadruples when [SO3] doubles (with H2O remaining constant) between Trials I & II. The order with respect to H2O is 1, as rate triples as [H2O] triples (with SO3 constant) in Trials I & IV. X can be calculated by plugging the values for Trial III into the rate expression. First, you need to find k, which can be done by plugging the values of any of the trials in: --> Trial IV: 0.039 = k[0.1]^2[0.03] --> k= 130 M^-2*s^-1. Since we have k, we can now plug in the values from Trial III to solve for X. --> 0.234 = 130[X]^2[0.02] --> X = 0.3 M. B. The order of the rxn is the sum of the exponents in the rate expression: (2+1) = 3. C. This we already solved for in A, in order to solve for X. The rate constant, k= 130 M^-2*s^-1. D. Substitute 0.2 M for 0.1 M for [SO3]. --> rate = 130[0.2]^2[0.03] = 0.156 M/s.

Assign oxidation numbers to the atoms in the following reaction to determine the oxidized & reduced species & the oxidizing & reducing agents: SnCl2 + PbCl4 --> SnCl4 + PbCl2.

All of these are neutral, so the oxidation # of each compound must add to 0. In SnCl2, , with 2 Cl's present with an oxidation # of -1, Sn must be +2. Similarly, SnCl4 means Cl = -4 & Sn = -4. It works the exact same way for the others, meaning the order, from left to right, is: (+2, -2) + (+4, -4) --> (+4, -4) + (+2, -2). The oxidation # of Sn increases from +2 to +4, so it loses e- & is oxidized, making it the reducing agent. Since oxidation # of Pb has decreased from +4 to +2, it gains e-, meaning it has been reduced, making it the oxidizing agent.

All of the following are true statements concerning catalysts except that a: A. Catalyst will speed up the rate-determining step. B. Catalysts will be used up in a reaction. C. Catalysts may induce steric strain on a molecule. D. Catalysts will lower the activation energy of a reaction.

B.

All of the following are true statements concerning reaction orders except for: A. The rate of a zero-order reaction is constant. B. After 3 ½-lives, a sample will have 1/9 of its original activity. C. The units for the rate constant for first-order reactions are s^-1. D. Higher-order reactions are those with an order greater than two.

B.

In the diagram provided, which labeled arrow represents the activation energy for the reverse reaction?

B. Ea is minimum amount of E needed for a rxn to proceed. The Ea for the ← rxn is the ΔEa between products & the transition state indicated by B.

Which of the following represents the net ionic equation for the reaction given? [AgNO3] (aq) + [Cu] (s) --> [CuNO3] (aq) + [Ag] (s). A. [AgNO3] + [Cu] --> [CuNO3] + [Ag]. B. [Ag+] + [Cu] --> [Cu+] + [Ag]. C. [AgNO3] + [Cu] --> [Cu+] + [NO3-] + [Ag]. D. [Ag+] + [NO3-] + [Cu] --> [CuNO3] + [Ag].

B. In displacement rxns, there are often ionic species that remain spectators in solution. Therefore, they can be written in terms of net ionic equations, which express the rxns of participating species. In this rxn, nitrate is a spectator ion, so the remainder are written out.

Which of the following choices correctly describes the solubility behavior of Potassium Chloride? A. Solubility in CCl4 > Solubility in CH3CH2OH > Solubility in H2O. B. Solubility in H2O > Solubility in CH3CH2OH > Solubility in CCl4. C. Solubility in CH3CH2OH > Solubility in CCl4 > Solubility in H2O. D. Solubility in H2O > Solubility in CCl4 > Solubility in CH3CH2OH.

B. KCl is an ionic salt, so soluble in polar solvents, insoluble in nonpolar solvents. Water is highly polar, so it would be first. The carbon atom in CCl4 is bonded to four atoms, meaning tetrahedral, cancelling out any polarity, meaning it would be last. Ethanol is somewhat polar, so it would fall in the middle.

Which of the following will be the most electrically conductive? A. Sugar dissolved in water. B. Salt water. C. Salt dissolved in an organic solvent. D. An oil & water mixture.

B. Only ionic compounds (electrolytes) dissolved in polar solvents will conduct electricity. Sugar is a covalent solid, & therefore not an electrolyte, even when dissolved in water. C & D are incorrect because salt will not dissolve appreciably in an organic solvent, & oil & water are immiscible. NaCl is an organic compound, so B is correct.

Which of the following is most likely to increase the rate of a reaction? A. Decreasing the temperature. B. Increasing the volume of the reaction vessel. C. Reducing the activation energy. D. Decreasing the concentration of the reactant in the reaction vessel.

C. Choice C is the only one that would INCREASE a rate of rxn, since it would need less Ea to proceed.

To what volume must 10.0 mL of 5.0 M HCl be diluted to make a 0.50 M HCl solution? A. 1.00 mL. B. 50.0 mL. C. 100 mL. D. 500 mL.

C. When a solution is diluted, more solvent is added, yet # of moles of solute remains the same. To solve a dilution problem, the following equation is used: --> M1V1 = M2V2. Therefore: --> (5.0 M)(10 mL) = (0.50 M)(x mL) --> x = 100 mL.

According to chemical kinetic theory, a reaction can occur: A. If the reactants collide with the proper orientation. B. If the reactants possess sufficient energy of collision. C. If the reactants are able to form a correct transition state. D. All of the above.

D.

What is the normality of a 2.0 M solution of phosphoric acid, H3PO4, for an acid-base titration? A. 0.67. B. 2.0. C. 3.0. D. 6.0.

D. Each mole of H3PO4 contains 3 moles of H &, since this is an acid, three mole equivalents. A 2 M solution of this acid is thus: --> 2 M x [3 N/1 M] = 6 N.

A simple cake icing can be made by dissolving a large quantity of sucrose in boiling water, cooling the mixture, & applying it to the cake before it reaches room T. The mixture hardens as it cools because: A. Sugar molecules freeze from liquid to solid. B. Common ion effect limits the dissolution of sucrose. C. Ksp of sucrose increases as the solution cools. D. Sugar concentration increases as water boils off, & solubility of sugar in water decreases as the solution cools.

D. Sugar-water solution hardens because it becomes supersaturated as it cools, since solubility of sucrose in water decreases significantly with temperature. Evaporation of water during boiling also contributes to the solution's supersaturated state by reducing the amount of solvent present. Choice A is incorrect, since sugar is dissolving, not melting.

What are the % compositions of Ag, N, & H in Ag(NH3)2^+? Ag(NH3)2^+ --> Ag^+ + 2NH3

Divide the mass of the element by total formula mass of compound. Mass of Ag(NH3)2^+ would be: (107.87) + (14.01∙2) + (1∙6) = 142 g/mol. Now / the individual molecular masses to the total mass to get % composition. % Ag = [108/142] ∙ 100 = 76.1%. % N= [(2*14)/142] ∙ 100 = 19.7%. % H = [(1*6)/142] ∙ 100 = 4.2%.

If 28 g of Fe are reacted with 24 g of S to produce FeS, what is the limiting reagent? How many grams of excess reactant are present in the vessel at the end of the reaction?

First, balance the equation. Fe + S --> FeS. Then, # of moles must be determined. 28 g Fe∙[1 mol Fe/56 g Fe] = 0.5 mol Fe. 24 g S∙[1 mol S/ 32 g S] = 0.75 mol S. Since 1 mole Fe is needed to react with 1 mol S per the equation, & 0.5 moles Fe for every 0.75 moles S we calculated, the limiting reagent is Fe. Thus, 0.5 moles Fe will react with 0.5 moles S, leaving an excess of 0.25 moles S in the vessel. mass of S = (0.25 mol S/1) ∙ (32 g/1 mol S) = 8 grams S.

What is the sum of the coefficients of the following equation once balanced? C6H12O6 + O2 --> CO2 + H2O

First, balance the equation. C6H12O6 + 6O2 --> 6CO2 + 6H2O. Then, count coefficients. 1 + 6 + 6 + 6 = 19.

How many grams of calcium chloride are needed to prepare 72 grams of silver chloride according to the following equation: CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl2.

First, determine if the equation is balanced. It is. 1 mol CaCl2 yields 2 mol AgCl2 when reacted with 2 mol AgNO3. Start with the goal mass of 72 grams, convert to moles, then find # of moles of CaCl2 required, then convert into g as the question specifies. [72 g AgCl/1] ∙ [1 mol AgCl/144 g AgCl] ∙ [1 mol CaCl/2 mol AgCl] ∙ [110 g CaCl2/1 mol CaCl2] = 27.5 g of CaCl2.

Determine the molecular formula & calculate % composition of each element present in nicotine, with an empirical formula of C5H7N, & a molecular mass of 162 g/mol.

First, determine molecular mass. 5(C) + 7(H) + 1(N) = empirical mass. 5(12 g/mol) + 7(1 g/mol) + 14 g/mol = 81 g/mol. Divide this by empirical mass to determine the # to multiply the subscript by. (162 g/mol) / (81 g/mol) = 2. 2(C5H7N) = C10H14N2. This is the molecular formula of Nicotine. Now we can find the % composition / molar mass by molecular mass of nicotine. % C = [(10∙12) / 162] ∙ 100 = 74.1%. % H = [(14∙1) / 162] ∙ 100 = 8.6%. % N = [14∙2) / 162] ∙ 100 = 17.3 %. Note: we would get the same answer using the empirical formula.

What is the empirical formula of a compound that contains 40.9% C, 4.58% H, & 54.2% O by mass with a molecular mass of 264 g/mol?

First, determine moles of each element by assuming a 100-g sample. # mol C = [40.9 / 12 g/mol] = 3.41 mol. # mol H = [4.58 / 1 g/mol] = 4.58 mol. # mol O = [54.2 / 16 m/mol] = 3.41 mol. Then, find simplest whole # ratio by / # of moles by smallest # from the previous step. C: 3.41/3.41 = 1. H: 4.58/3.41 = 1.3. O: 3.41/3.41 = 1. This leads us to C1 H1.3 O1. Now, multiply by 3 in order to obtain all #s. The answer then becomes: C3H4O3.

How many moles are in 9.52 g of MgCl2?

First, find molecular mass of MgCl2. --> (24.31) + (35.45∙2) = 95.21 g/mol. Second, / grams by grams per mole to result in just moles. --> 9.52 g/95.21 g/mol = 0.10 mol MgCl2.

What is the % yield for a reaction in which 27 g of Cu is produced by reacting 32.5 g Zn in excess CuSO4 solution?

First, find the balanced equation. Zn + CuSO4 --> Cu + ZnSO4. Now, calculate the theoretical yield for Cu. 32.5 g Zn ∙ [1 mol Zn/65 g Zn] = 0.5 mol Zn. 0.5 mol Zn ∙ [1 mol Cu/1 mol Zn] = 0.5 mol Cu. 0.5 mol Cu / [64 g Cu/1 mol Cu] = 32 g Cu. This is the theoretical yield. Now find % yield. [27 g / 32 g] ∙ 100 = 84%.

What is the % Composition of chromium in K2Cr2O7?

First, find the total mass. --> (39.10∙2) + (52∙2) + (16∙7) = 294.2 g/mol. Then, use the % composition equation. --> [(2(52 g/mol))/294 g/mol) ∙ 100] = 35.4 %.

Consider the following hypothetical reaction & experimental data: A + B --> C + D T = 273 K Trial I: [A]: 0.10 mol/L. [B]: 1 mol/L. Rate: 0.035 mol/L*s. Trial II: [A]: 0.10 mol/L. [B]: 4 mol/L. Rate: 0.070 mol/L*s. Trial III: [A]: 0.20 mol/L. [B]: 1 mol/L. Rate: 0.140 mol/L*s. Trial IV: [A]: 0.10 mol/L. [B]: 16 mol/L. Rate: 0.140 mol/L*s. A. What is the order with respect to A? B. What is the order with respect to B? C. What is the rate equation? D. What is the overall order of the reaction? E. Calculate the rate constant.

First, general rate equation must be written out. k[A]^x[B]^y. k= rate constant. x= order with respect to [A]. y= order with respect to [B]. A. x = 2. To solve for x, find 2 trials where B is constant, so 1 & 3. If [A] is doubled, the rate ↑ by a factor of 4 (0.140/0.035). Thus, rate varies as the square of [A]. The order for [A] is therefore 2. B. y = 0.5. To solve for y, find 2 trials where A is constant, so 1 & 2. [B] quadruples, while rate doubles. The rate, therefore, varies as the square root of [B]. So the order for [B] is 0.5. C. Rate = k[A]^2[B]^0.5. D. Order = 2.5. The overall order = x + y = 2.5. E. 3.5. Given the rate expression, the rate constant can be calculated by substituting the rate & concentrations for any of the 4 trials into the rate expression. The rate comes out to 3.5 in each case: Trial I: 0.035 = k[0.10]^2[1]^0.5 --> 3.5 M^1.5 s^-1. Trial II: 0.070 = k[0.10]^2[4]^0.5 --> 3.5 M^1.5 s^-1. Trial III: 0.140 = k[0.20]^2[1]^0.5 --> 3.5 M^1.5 s^-1. Trial IV: 0.140 = k[0.10]^2[16]^0.5 --> 3.5 M^1.5 s^-1.

If enough water is added to 11 g of CaCl2 to make 100 mL of solution, what is the molarity of the solution? Note: (molecular mass of CaCl2 is 110 g/mol).

First, we need CaCl2 in moles. --> [11 g CaCl2/110 g/mol CaCl2] = 0.10 mol CaCl2. Now, we need to convert mL to L. --> [100 mL/1] ∙ [1 L/1,000 mL] = 0.10 L. Molarity = moles solute/L solution, therefore: --> [0.10 mol CaCl2/0.10 L solution] = 1.0 M CaCl2.

If 10 g of NaOH are dissolved in 500 g of H2O, what is the molality of the solution? Note: (the molecular mass of NaOH of is 40 g/mol).

First, we need moles of NaOH. --> [10 g NaOH/1] ∙ [1 mol NaOH/40 g NaOH] = 0.25 mol of NaOH. Then, we need kilograms of water. --> [500 g H2O/1] ∙ [1 kg/1,000 g] = 0.50 kg H2O. Now, we can solve the equation. --> m = [mol solute/kg solvent] = [0.25 mol NaOH/0.50 kg H2O] = 0.50 m NaOH.

Assign oxidation numbers to each atom of the following reaction: 2 Fe(s) + O2(g) + 2H2O(l) --> 2Fe(OH)2 (s).

From Left to Right: Fe: 0. O: 0. H: +1. O: -2. Fe: +2. O: -2. H: +1.

The number of undecayed nuclei in a sample of Bromine-87 decreased by a factor of 4 over a period of 112 s. What is the decay constant for Bromine-87? A. 6.19 x 10^-3 s^-1. B. 1.24 x 10^-2 s^-1. C. 6.93 x 10^-1 s^-1. D. 56 s.

If the # of nuclei decaying in a sample has decreased by a factor of 4, the sample has been through 2 ½-lives, & so the ½-life will be: --> [(112 s)/(2)] = 56 s = t½. The equation to determine the decay constant for a 1-order rxn is: --> t½ = 0.693/k --> 56 seconds = 0.693/k. --> k = 0.0124 s^-1.

A hydrocarbon is heated in an excess of oxygen to produce 58.67 g of CO2 & 27 g H2O. What is the empirical formula of the hydrocarbon?

Let's start with what we know. [(58.67 g CO2) / (44 g/mol CO2)] = 1.33 mol CO2. [(27 g H2O) / (18 g/mol H2O)] = 1.5 mol H2O. In the first part, we have 1 mol of C in CO2, so it's 1.33 mol. In the second part, we have 2 mol of H in H2O, so it's 1.5(2) = 3 mol H. Now we have C1.33H3. Since we cannot keep it in decimal form, we can use the following: 3(C1.33H3) = C4H9.

If 5.8 g of Ag(NH3)2^+ yields 1.4 g of NH3, how many moles of silver are produced? Ag(NH3)2^+ --> Ag^+ + 2NH3

Since mass of products must always be = on both sides, we can infer that Silver Nitrate = Ammonia + Silver, therefore: --> 5.8 g = 1.4 g + x --> x = 4.4 g Ag. Then we just / by Ag molecular mass to find # of moles. 4.4 g Ag/ 108 g/mol = 0.041 mol Ag.

The percent composition of an unknown element X in CH3X is 32%. Which of the following element is X? A. H B. F C. Cl D. Li

Since we know that x is 32%, we know that CH3 is 68%. We can calculate the formula mass of CH3: --> 12 + 1(3) = 15 g. Since 15 g = 68%, we can then solve for the remainder: 15 grams is 68% of x. x = 22 grams. 22 grams - 15 grams = 7 grams. Then, we just have to figure out which element has a molecular mass of 7. The answer must be D.

Balance the following reaction: NF3 + H2O --> HF + NO + NO2. Then, how many grams of HF are expected to form if 1.5 kg of a 5.2% NF3 sample is used?

The balanced equation will be: 2 NF3 + 3 H2O --> 6 HF + NO + NO2. 2 mol NF3 produces 6 mol HF. First, we have to determine the amount of NF3 from the equation. 1,500 g ∙ 0.052 = 78 g NF3. Now, we can use basic stoichiometry. [78 g NF3/1]∙[1 mol NF3/71 g NF3]∙[6 mol HF/2 mol NF3]∙[20 g HF/1 mol HF] = 66 g HF.

What is the mass in grams of single chlorine atom? Of a single molecule of O2?

The mass of a single atom is determined by dividing the atomic mass by Avogadro's number. Cl: [(35.5 g/mol)/(6.022 ∙ 10^-23)] = 5.89 ∙ 10^-23 g/atom. O2: [((16∙2)g/mol)/(6.022∙10^-23)] = 5.31 ∙ 10^-23 g/molecule.

Balance the redox reaction between MnO4 & I in a basic solution: MnO4- + I- --> I2 + Mn2+.

The oxidation #s for I & Mn change: I changes from -1 to 0, & Mn from +7 to +2, so the rxn is not balanced. So, use the 5-step method. Step I: Separate the 2 ½-rxns. --> [I-] --> [I2]. --> [MnO4-] --> [Mn2+]. Step II: Balance atoms of ½-rxns. 1st, balance all atoms except H & O. Next, add H2O to balance O atoms & add H+ to balance H atoms. --> [2I-] --> [I2]. The Mn in permanganate is already balanced, so balance O atoms by adding 4 H2O to the right side. --> [MnO4-] --> [Mn2+ + 4H2O]. Finally, add H+ to the left side to balance the 4 H2Os. --> [MnO4-] + [8H+] --> [Mn2+] + [4H2O]. Step III: Balance the e- on each ½-rxn. For I2-, since it is 0, we must add e-. --> [2I-] --> [I2] + [2 e-]. For Mn, there is a charge of +7 on left, & +2 on right. --> [MnO4] + [8H+] + [5 e-] --> [Mn2+] + [4H2O]. Next, both ½-rxns must have the same # of e- when combined, so multiply the oxidation ½ by 5, & the reduction ½ by 2. Oxidation ½: --> 5([2I-] --> [I2] + [2 e-]). --> [10 I-] --> [5I2] + [10 e-]. Reduction ½: --> 2([MnO4-] + [8H+] + [5 e-] --> [Mn2+] + [4H2O]). --> [2MnO4-] + [16H+] + [10 e-] --> [2Mn2+] + [8H2O]. Step IV: Combine ½-Rxns. [10 I-] + [10 e-] + [16 H+] + [2MnO4-] --> [5 I2] + [10 e-] +[2Mn2+] + [8H2O]. Then, add [16 OH-] to both sides to neutralize [H+]: --> [10 I-] + [10 e-] + [16 H+] + [2MnO4-] + [16OH-] --> [5 I2] + [10 e-] +[2Mn2+] + [8H2O] + [16OH-]. Now, combine extra [H+] & [OH-] into [H2O]. --> [10 I-] + [10 e-] + [16H2O] + [2MnO4-] --> [5I2] + [10 e-] + [2Mn2+] + [8H2O] + [16OH-]. Now, cancel out any duplicate species (H2O & e-). --> [10 I-] + [2MnO4-] --> [5I2] + [2Mn2+] + [16OH-]. Step V: Check that everything is balanced.

Balance the redox reaction between MnO4 & I in an acidic solution: MnO4- + I- --> I2 + Mn2+.

The oxidation #s for I & Mn change: I changes from -1 to 0, & Mn from +7 to +2, so the rxn is not balanced. So, use the 5-step method. Step I: Separate the 2 ½-rxns. --> [I-] --> [I2]. --> [MnO4-] --> [Mn2+]. Step II: Balance atoms of ½-rxns. 1st, balance all atoms except H & O. Next, add H2O to balance O atoms & add H+ to balance H atoms. --> [2I-] --> [I2]. The Mn in permanganate is already balanced, so balance O atoms by adding 4 H2O to the right side. --> [MnO4-] --> [Mn2+ + 4H2O]. Finally, add H+ to the left side to balance the 4 H2Os. --> [MnO4-] + [8H+] --> [Mn2+] + [4H2O]. Step III: Balance the e- on each ½-rxn. For I2-, since it is 0, we must add e-. --> [2I-] --> [I2] + [2 e-]. For Mn, there is a charge of +7 on left, & +2 on right. --> [MnO4] + [8H+] + [5 e-] --> [Mn2+] + [4H2O]. Next, both ½-rxns must have the same # of e- when combined, so multiply the oxidation ½ by 5, & the reduction ½ by 2. Oxidation ½: --> 5([2I-] --> [I2] + [2 e-]). --> [10 I-] --> [5I2] + [10 e-]. Reduction ½: --> 2([MnO4-] + [8H+] + [5 e-] --> [Mn2+] + [4H2O]). --> [2MnO4-] + [16H+] + [10 e-] --> [2Mn2+] + [8H2O]. Step IV: Combine ½-Rxns. [10 I-] + [10 e-] + [16 H+] + [2MnO4-] --> [5 I2] + [10 e-] +[2Mn2+] + [8H2O]. To find the final equation, cancel out e- & anything that appears on both sides of the equation, as those would just be spectators. --> [10 I-] + [16H+] + [2MnO4-] --> [5 I2] + [2Mn2+] + [8H2O]. Step V: Confirm that mass & charge are balanced. Here, there is a +4 charge on each side, & atoms are stoichiometrically balanced. This is a balanced equation.

A hydrogen-filled balloon was ignited & 1.50 g of H reacted with 12.0 g of O. How many grams of H2O vapor formed? (Assume that water vapor is the only product).

Tricky question! This can be done mathematically, however, according to the law of conservation of mass: 1.50 g + 12.0 g = 13.5 g of H2O formed.

How many mL of a 5.5 M NaOH solution must be used to prepare 300 mL of a 1.2 M NaOH solution?

Use M1V1 = M2V2. --> 5.5 M ∙ V1 = 1.2 M ∙ 300 mL. --> V1 = [1.2 M ∙ 300 mL]/[5.5 M]. --> V1 = 65 mL.

CaC2 + 2H2O --> Ca(OH)2 + C2H2. How many grams of C2H2 are formed from 0.400 moles of CaC2?

Using the equation, 1 mol C2H2 is formed when it reacts with 1 mol CaC2. [0.4 mol CaC2 / 1] ∙ [1 mol C2H2/1 mol CaC2] ∙ [26 g C2H2 / 1 mol C2H2] = 10.4 g C2H2.

27 g of Ag was reacted with excess S, according to the following equation: 2Ag + S --> Ag2S. 25.0 g of Ag2S was collected. What are the theoretical yield, actual yield, & % yield?

We already have our actual yield, which was 25.0 grams. We can calculate the theoretical yield using the molecular formula given. [27 g Ag/1] ∙ [1 mol Ag/108 g Ag] ∙ [1 mol Ag2S/2 mol Ag] ∙ [248 g Ag2S/1 mol Ag2S] = 31 g Ag2S. Now that we have both theoretical & actual yield, we can find % yield. [25.0 g/31.0 g] ∙ 100 = 81%.

Calculate the Molecular Mass of SOCl2.

We have to add every atom up based on the periodic table. S = 32.07 amu. O = 16.00 amu. Cl = 35.45 amu. It's important to take into consideration that there are two Cl, as denoted by the subscript. Therefore: (32.07) + (16.00) + (35.45∙2) = 118.97 or 119 amu.

What is the molecular formula of a compound that contains 40.9% C, 4.58% H, & 54.2% O by mass with a molecular mass of 264 g/mol?

When molecular mass is given, this is easy. # mol C = (0.409)(264) / 12 g/mol = 9. # mol H = (0.0458)(264) / 1 g/mol = 12 mol. # mol O = (0.542)(264) / 16 g/mol = 9 mol. Therefore, its C9H12O9.

What is the percent composition by mass of NaCl of a saltwater solution if 100 grams of solution contains 20 grams of NaCl?

[ 20 grams NaCl/100 g ] ∙ 100 = 20% NaCl solution.

How many moles of Ag(NH3)2^+ are required for the production of 11 moles of ammonia using the following equation: Ag(NH3)2^+ --> Ag^+ + 2NH3

[11 mol NH3/1]∙[1 mol Ag(NH3)2^+/2 mol NH3] = 5.5 mol Ag(NH3)2^+.

Using the ion-electron method, balance the following equation of a reaction taking place in an acidic solution: [ClO3-] + [AsO2-] --> [AsO4^3-] + [Cl-].

[ClO3-] + [3H2O] + [3AsO2-] --> [3AsO4^3-] + [Cl-] + [6H+].

Find the rate law for data below for the following reaction at 300 K. A + B --> C + D. Trial 1: [A]initial = 1.00 M. [B]initial= 1.00 M. rinital= 2.0 M/S. Trial 2: [A]initial = 1.00 M. [B]initial= 2.00 M. rinital= 8.1 M/S. Trial 3: [A]initial = 2.00 M. [B]initial= 2.00 M. rinital= 15.9 M/S.

a. In Trials 1 & 2, the conc. of A is kept constant, while B is doubled. The rate increases by a factor of 8.1/2.0, or ~4. Write down the rate expression for each. Trial 1: r1= k[A]^x k[B]^y = k(1.00)^x(1.00)^y. Trial 2: r2= k[A]^x k[B]^y = k(1.00)^x(2.00)^y. / the 2nd equation by the 1st. r2/r1= [8.1/2] = [k(1.00)^x(2.00)^y/k(1.00)^x(1.00)^y]. 4 = (2.00)^y --> y= 2. b. In Trials 2 & 3, the conc. of B is kept constant, while A is doubled; rate increases by a factor of 15.9/8.1, or ~2. Rate expressions for each are: Trial 2: r2= k[A]^x k[B]^y = k(1.00)^x(2.00)^y. Trial 3: r3= k[A]^x k[B]^y = k(2.00)^x(2.00)^y. / 2nd equation by the 1st. r3/r2= [15.9/8.1] = [k(2.00)^x(2.00)^y/k(1.00)^x(2.00)^y]. 2 = (2.00)^x --> x= 1. So, r = k[A][B]^2. The overall rxn order is 1 + 2 = 3. To calculate k, substitute values from anyone of the trials into the rate law. 2.0 M/S = k (1.00 M)(1.00 M)^2. k = 2.0 M^-2 s^-1. Therefore, the rate law is r= (2.0 M^-2 s^-1)[A][B]^2.