Chemistry 161 midterm 2

i = insoluble

( < 0.1 g solute/100 g of water )

s = soluble

( > ~ 1 g solute/100 g of water )

d = decomposes

( in water )

ss = slightly soluble

( ~ 0.1−1 g solute/100 g of water )

A 50.0 mL portion of a 1.70 M solution is diluted to a total volume of 268 mL. A 134 mL portion of that solution is diluted by adding 137 mL of water. What is the final concentration? Assume the volumes are additive.

(1.70 M)(50.0 mL1000 mL/L)=0.0850 mol Note that the number of moles does not change during dilution, so 0.0850 mol is the amount of solute present in the 50.0‑mL solution as well as in the 268‑mL solution. Since 134 mL is one‑half of 268 mL, the 134‑mL portion will contain one‑half the number of moles as the 268‑mL solution. 134 mL268 mL(0.0850 mol)=12(0.0850 mol)=0.0425 mol After 137 mL is added to the 134‑mL portion, the final volume is 137+134=271 mL. The final 271‑mL solution contains 0.0425 mol of solute and so the concentration is 0.0425 mol271 mL×1 L1000 mL=0.157 M

Write the equations for the pairs that will react. Al(s) + SnCl2(aq) → 2 AlCl3(aq) + 3 Sn(s)

(Al is more active than Sn)

Write the equations for the pairs that will react. Li(s) + Cr(s) → no reaction

(BOTH ARE METALs)

Determine whether the following reactants will react with HCl(aq) and write balanced equations. Cu(s) + HCl(aq) → no reaction

(Cu is less active than H)

Write the equations for the pairs that will react. Zn2+(aq) + Mg2+(aq) → no reaction

(Mg is more active than Zn)

Determine whether the following reactants will react with HCl(aq) and write balanced equations. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

(Zn is more active than H)

Determine the oxidation state of each species: Identify the oxidation state of Ba2+.

+2

Determine the oxidation state of each species: Identify the oxidation state of Zn in ZnSO4.

+2

Determine the oxidation state of Cl in each species: Identify the oxidation state of Cl in Ba(ClO2)2.

+3

Determine the oxidation state of each species: Identify the oxidation state of S in SO2

+4

Determine the oxidation state of Cl in each species: Identify the oxidation state of Cl in ClF+4.

+5

Determine the oxidation state of each species: Identify the oxidation state of S in SO2−4.

+6

Determine the oxidation state of Cl in each species: Identify the oxidation state of Cl in Cl2O7.

+7

Titration

- Titration is a laboratory technique for determining the number of moles of a substance dissolved in an aqueous solution (sample solution). - One reactant, the standard solution, has a precisely known concentration. - The volume of the standard solution and the sample are both carefully measured. - The mole ratios and volumes are used to calculate the sample concentration.

Determine the oxidation state of Cl in each species: Identify the oxidation state of Cl in AlCl−4.

-1

Weak Acids are Weak Electrolytes

-All acids are weak except for the seven strong acids. -Weak acids dissolve as molecules but do not dissociate well and therefore only a small percentage of ions form. -Weak acids form solutions that conduct electricity only slightly and are weak electrolytes.

Compounds in Aqueous Solution

-Aqueous implies water. - A solution implies a solvent containing a solute. -A solid compound that readily dissolves in water is soluble in water. -A compound that remains a solid in water is said to be insoluble in water. -The notation NaCl(aq) implies that NaCl is has dissolved in water. -In a reaction of aqueous substances, a notation (s) for the product AgCl(s) indicates that AgCl has precipitated.

Balancing Equations

-Chemical reactions obey the law of conservation of mass. -All atoms present at the beginning of a reaction must be present at the end of the reaction. -In a balanced equation, the number of atoms of each element on the left side of the balanced equation is equal to the number of atoms of that element on the right.

Standard Enthalpies of Formation

-Determine the enthalpy of formation of a chemical compound from its constituent elements in their standard states. -The standard state of a substance is the state in which the substance is most stable at 25°C (298 K) and 1 atm pressure. -The standard enthalpy of formation, ΔH° , is the enthalpy of formation of 1 mol of a 𝒇 compound from its constituent elements in their standard states (25°C and 1 atm). H2(g) + O2(g) → 2 H2O(l) -Note that the ΔH° of any element in its standard state is 0.

Review: Determine Precipitate Formation

-Determine the two new ionic compounds that could form. Write their formulas based on the charges of the ions. -Use the solubility chart (guidelines) to determine if either possible product is insoluble. • If both possible products are soluble, no reaction will occur.• If one product is insoluble, a precipitation reaction will occur.

Energy and Energy Units

-Energy is the capacity to do work or transfer heat. -Mechanical energy is due to an object's motion, position, or both. -Kinetic energy is energy of motion. KE=1𝑚𝑣2 2 -Potential energy is energy related to position.

Example: Liquid ethanol at 25°C

-Ethanol(l) has a specific internal energy, Uliquid. -Ethanol(l) could have been formed by melting solid ethanol or by condensing ethanol vapor. -In either case, the internal energy of the liquid ethanol is the same. (Temperature is a state function) -Yet, the heat or work associated with arriving at liquid ethanol is different in the two cases. (w and q path functions)

Heat of Solution

-For processes happening in solution, 𝒒𝐬𝐨𝐥𝐧 = −𝒒𝐩. -The release or absorption of heat that occurs when a solid compound dissolves in a solvent at constant pressure to infinite dilution is known as the enthalpy solution, ΔHsoln, or heat of solution. -Both reaction enthalpies and heats of solution are commonly measured using constant-pressure calorimetry.

Weak electrolyte

-HC2H3O2 -NH3

Seven Strong Acids (all others are weak)

-Hydrochloric acid HCl H+(aq) + Cl-(aq) - Hydrobromic acid HBr H+(aq) + Br-(aq) - Hydroiodic acid HI H+(aq) + I-(aq) -Nitric acid HNO3 H+(aq) + NO3-(aq) -Perchloric acid HClO4 H+(aq) + ClO4-(aq) -Chloric Acid HClO3 H+(aq) + ClO3-(aq) -Sulfuric acid H2SO4 H+(aq) + HSO4-(aq)

Seven Strong Acids

-Hydrochloric acid HCl H+(aq) + Cl-(aq) -Hydrobromic acid HBr H+(aq) + Br-(aq) -Hydroiodic acid HI H+(aq) + I-(aq) -Nitric acid HNO3 H+(aq) + NO3-(aq) -Perchloric acid HClO4 H+(aq) + ClO4-(aq) -Chloric Acid HClO3 H+(aq) + ClO3-(aq) -Sulfuric acid H2SO4 H+(aq) + HSO4-(aq)

Double-Replacement Reactions

-In a double-replacement reaction (also known as a double displacement or metathesis reaction), involves two ionic compounds that exchange ions to form two new compounds. -Lead(II) nitrate reacts with potassium iodide in a double-replacement reaction. 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)

A reaction occurring in a flask releases 890 J of heat to the surroundings and the gas produced performs 450 J of work on the surroundings by pushing the piston upward.

-Is this a closed, open, or isolated system? Closed (energy but not matter) -Determine the signs of q and w in this reaction. q < 0 and w < 0 -Determine ΔU for this reaction. 𝚫𝑼 = 𝒒 + 𝒘𝜟𝑼 = (−𝟖𝟗𝟎 𝐉) + (−𝟒𝟓𝟎 𝐉) 𝜟𝑼 = −𝟏𝟑𝟒𝟎 𝐉

Enthalpy in Chemical Reactions

-It is most often given after a chemical equation indicating the enthalpy changed associated with the reaction. -Stoichiometry applies.

Strong electrolyte

-NaCl -HCl -NaOH

Calculations with Net Ionic Equations

-Net ionic equations, like all other balanced chemical equations, give the mole ratios of reactants and products. -However, the mass of the appropriate spectator ions must be included in mass calculations. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) Ag+(aq) + Cl−(aq) → AgCl(s) -We can calculate the amount of Ag+ needed to make 100 g of AgCl. But, we cannot weigh out Ag+ without including its counter ion.

Net Ionic Equations

-Some ions are unchanged. • These spectator ions do not participate in the reaction.• Removing the spectator ions leaves the net ionic equation: Ag+(aq) + Cl-(aq) → AgCl(s)

Molarities of Ions

-Strong electrolytes dissociate into ions when dissolved in water. -The molarity of any ion is the number of moles of that ion per liter of solution. -The subscripts in a chemical formula indicate the mole ratio of the compound to its constituent elements -In a 1.0 M solution of AlCl3(aq), the ion concentrations are 1.0 M Al3+ and 3.0 M Cl-.

Hess's Law

-The enthalpy changes for some reactions are difficult to measure directly. -These can be calculated by manipulating the equations for other reactions with known ΔH values. -When chemical equations are added together to yield a different chemical equation, the corresponding ΔH values are added to get the ΔH for the desired equation. (This principle is known as Hess's law.)

Specific Heat

-The heat required to change the temperature of a substance. -In the absence of a phase change, the amount of heat required to change the temperature of a substance is given by: 𝒒 = 𝒎𝒄𝚫𝑻 -...where m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature in either degrees Celsius or kelvins. -The specific heat, c, of a substance is the number of calories required to raise exactly 1 g of the substance by exactly 1°C. Typical units are J/(g∙°C). 𝒒 = 𝒎𝒄𝚫𝑻

Energy Units (Joule vs Calorie)

-The joule, J, is the SI unit of energy. 𝟏 𝐉=𝟏 𝐤𝐠·𝐦𝟐/s𝟐 -The calorie, Cal, is another commonly used unit of energy. -A calorie is the amount of energy needed to raise the temperature of exactly 1 g of water by exactly 1°C. 𝟏 𝐜𝐚𝐥 = 𝟒.𝟏𝟖𝟒 𝐉 𝐞𝐱𝐚𝐜𝐭𝐥𝐲

Predicting the Products of Redox Reactions

-The reactants are less stable, more reactive, than the products. -Ions are said to be reactive. -Active metals oxidize (readily give up electrons). -Active metals transfer electrons to ions of less active metals.

Consider the reaction shown. 4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.125 mol HCl reacts with an excess of O2.

0.125 mol HCl×2 mol Cl24 mol HCl=0.0625 mol Cl2 Then, convert the moles of Cl2 formed to grams using the molar mass of Cl2. 0.0625 mol Cl2×70.906 g Cl21 mol Cl2=4.43 g Cl2 The two-step calculation can be combined into one calculation. 0.125 mol HCl×2mol Cl24mol HCl×70.906 g Cl21mol Cl2=4.43 g Cl2

Consider the reaction. Pb(SO4)2+2Zn⟶2ZnSO4+Pb If 0.592 mol of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate will be produced in the reaction?

0.592 mol Zn X 2 mol ZnSO4 / 2 mol Zn X 161.47 g ZnSO4 / 1 mol ZnSO4 = 95.6 g ZnSO4

Steps for balancing a chemical equation:

1 .Balance polyatomic ions first if they are present as both reactants and products. For example, begin with sulfate SO42− when it appears on both sides of the arrow (do not attempt to balance S and O separately). 2. Balance elements that appear in only one reactant and one product, reserving any that appear in more than one reactant or more than one product until last. 3. Balance any remaining elements. This may require changing previously determined coefficients. 4. Verify the balanced equation.

Balance the equation: Ba(OH)2 + HBr → BaBr2 + H2O

1 Ba(OH)2 + 2 HBr → 1 BaBr2 + 2 H2O

Balance the equation: KIO3 + KI + HCl → I2 + H2O + KCl

1 KIO3 + 5 KI + 6 HCl → 3 I2 + 3 H2O + 6 KCl

joule (J)

1 kg · m2/s2

Rules for Assigning Oxidation States

1. A neutral element that is not part of a compound has an oxidation state of zero. 2. Monoatomic ions have oxidation states equal to their ionic charges. 3. The sum of the oxidation states in any formula is equal to the overall charge on that formula. 4. Oxygen tends to have an oxidation state of -2 in compounds. 5. Hydrogen tends to have an oxidation state of +1 in compounds.

Steps in a Titration

1. One reactant, the titrant, is slowly added to the other.• The titrant is in a buret and the other reactant is in an Erlenmeyer flask to enable mixing by swirling. 2. An indicator is added to the reaction to indicate the end point of the titration. At the end point, the reaction is at (or very close to) the equivalence point, where all of the substance being titrated has reacted with the titrant. 3. The volume of the titrant added is determined by comparing the initial and final buret volumes.

Manipulating Chemical Equations Rules

1. When an equation is reversed, the sign of its enthalpy changes. 2. When the coefficients in an equation are multiplied or divided by a factor, the enthalpy value is multiplied or divided by that same factor. 3. When reactions are summed, the enthalpy of the overall reaction is the sum of the enthalpies of the component reactions.

Calorie (Cal)

1000cal = 1kcal = 4184J = 4.184kJ

Write the balanced equation for the reaction of sodium hydroxide with phosphoric acid to produce sodium hydrogen phosphate, Na2HPO4, and water.

2 NaOH + 1 H3PO4 → 1 Na2HPO4 + 2 H2O

For the chemical reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many moles of calcium carbonate (CaCO3) are produced from 4.0 mol of sodium carbonate (Na2CO3)?

4.0 mol Na2CO3 X 1 mol CaCO31 mol / 1Na2CO3 = 4.0 mol CaCO3

calorie (cal)

4.184 J

Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.120 L sample of an unknown HNO3 solution required 40.9 mL of 0.150 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution?

40.9 mL Ba(OH)2×1 L1000 mL=0.0409 L Ba(OH)2 0.150 molL Ba(OH)2×0.0409 L Ba(OH)2=0.00614 mol Ba(OH)2 Then, use the coefficients in the balanced chemical equation to relate the number of moles of Ba(OH)2 to the number of moles of HNO3. 0.00614 mol Ba(OH)2×2 mol HNO31 mol Ba(OH)2=0.0123 mol HNO3 Finally, determine the concentration of the HNO3 solution. molarity=moles of soluteliters of solution M=0.0123 mol HNO30.120 L HNO3=0.102 M HNO3

The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.40 mol of ethane is burned in an excess of oxygen?

5.40 mol C2H6 X 4 mol CO2 / 2 mol C2H6 = 10.8 mol CO2

Synthesis

A + B → AB

Single-Replacement

A + BC → AC + B

Double-Replacement

AB + CD → AD + CB

Acid-Base Reactions are Double-Replacement Reactions

AB + CD → AD + CB HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) acid + base → salt + water H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2 H2O(l) Note: An acid is usually written with H at the beginning of its formula.

Decomposition

AB → A + B

the best oxidizing agent.

Ag+

the most reactive metal ion.

Ag+

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) How many grams of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid? mass of CaCl2: Which reactant is in excess? How many grams of the excess reactant will remain after the reaction is complete?

Because the quantities of two reactants are given, you must check to see if one of them is a limiting reactant. The stoichiometric correspondences are 27.0 g CaCO3×1 mol CaCO3100.086 g CaCO3=0.270 mol CaCO3 10.0 g HCl×1 mol HCl36.461 g HCl=0.274 mol HCl Therefore, HCl is the limiting reactant because 0.274 mol HCl requires only 12(0.274 mol)=0.137 mol CaCO3, which is present in excess. The mass of calcium chloride produced is 0.274 mol HCl×1 mol CaCl22 mol HCl×110.98 g CaCl21 mol CaCl2=15.2 g CaCl2 Only 0.137 mol CaCO3 will react, so there is an excess (0.270−0.137) mol=0.133 mol. 0.133 mol CaCO3×100.086 g CaCO31 mol CaCO3=13.3 g CaCO3

Nonelectrolyte

C12H22O11

Determine whether the following reactants will react with HCl(aq) and write balanced equations. Ca2+(aq) + HCl(aq) → no reaction

Ca2+ is an ion (BOTH IONs)

An aqueous potassium carbonate solution is made by dissolving 7.05 moles of K2CO3 in sufficient water so that the final volume of the solution is 4.90 L. Calculate the molarity of the K2CO3 solution.

Calculate the molarity of the solution by dividing the moles of K2CO3 by the volume of the final solution in liters. molarity(M)=7.05 mol K2CO34.90 L=1.44 M

An aqueous magnesium chloride solution is made by dissolving 6.44 moles of MgCl2 in sufficient water so that the final volume of the solution is 4.80 L. Calculate the molarity of the MgCl2 solution.

Calculate the molarity of the solution by dividing the moles of MgCl2 by the volume of the final solution in liters. molarity(M)=6.44 mol MgCl24.80 L=1.34 M

What NaCl concentration results when 219 mL of a 0.650 M NaCl solution is mixed with 667 mL of a 0.340 M NaCl solution?

Calculate the moles of NaCl in each solution. 𝑛2=(0.340 M)(667 mL1000 mL/L)=0.227 mol 𝑛1=(0.650 M)(219 mL1000 mL/L)=0.142 mol Next, find the total number of moles and total volume. (219 mL)+(667 mL)=886 mL (0.142 mol)+(0.227 mol)=0.369 mol Finally, apply these totals to the molarity formula, being sure to include a conversion factor for volume. 𝑀=𝑛𝑉=0.369 mol886 mL×1 L1000 mL=0.417 M

H2CO3

Carbonic acid

Complete the equation for the dissociation of CdCl2(aq). Omit water from the equation because it is understood to be present.

CdCl2(aq)⟶Cd2+(aq)+2Cl−(aq)

Identify any redox processes. 2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)

Combustion redox

Solutions

Concentration is reported in molarity that can be used to calculate moles or volume

Combustion

CxHy + O2 → CO2 + H2O

CaCO3(s) → CaO(s) + CO2(g)

Decomposition reaction; oxidation-reduction is the driving force.

Reduction

Decrease (reduction) in oxidation state Gain of electrons Oxidizing agent

Identify any redox processes. MgI2(aq) + 2 AgNO3(aq) → 2 AgI(s) + Mg(NO3)2(aq)

Double replacement not redox

MgI2(aq) + 2 AgNO3(aq) → 2 AgI(s) + Mg(NO3)2(aq)

Double-replacement reaction; precipitation is the driving force.

Which solution contains the largest number of chloride ions? 100.0 mL of 0.80 M NaCl 50.0 mL of 0.60 M MgCl2 150 mL of 0.20 M AlCl3

Each salt dissociates in aqueous solution as shown. NaCl(s)⟶Na+(aq)+Cl−(aq) MgCl2(s)⟶Mg2+(aq)+2Cl−(aq) AlCl3(s)⟶Al3+(aq)+3Cl−(aq) One mole of NaCl will therefore produce one mole of chloride ions, whereas MgCl2 and AlCl3 will produce two and three moles, respectively. To calculate the moles of chloride ions in each solution, multiply the molarity of each solution by the volume in liters. Then, multiply by the number of chloride ions produced. The solution consisting of 150 mL of 0.20 M AlCl3 contains 0.090 mol Cl−. The solution consisting of 50.0 mL of 0.60 M MgCl2 contains 0.060 mol Cl−, and the solution consisting of 100.0 mL of 0.80 M NaCl contains 0.080 mol Cl−.

𝑯 = 𝑼 + 𝑷𝑽 𝚫𝑯 = 𝚫𝑼 + 𝑷𝚫𝑽

Enthalpy, H, is a state property that is related to internal energy, pressure, and volume by the following equations.

The oxidation state of Fe in FeCl3

Fe Cl3 x + 3(−1) = 0 x − 3= 0 x = +3

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 61.3 g of each reactant? 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

First, calculate the amount of H2O that can be produced from 61.3 g NH3 and excess O2. For NH3, the molar mass is 14.007 g/mol+3(1.0079g/mol)=17.0307 g/mol, and there are six mol H2O produced per four mol NH3 that react. For H2O, the molar mass is 2(1.0079g/mol)+15.999 g/mol=18.0148 g/mol. 61.3 g NH3×1 mol NH317.0307 g NH3×6 mol H2O4 mol NH3×18.0148 g H2O1 mol H2O=97.3 g H2O Next, calculate the amount of H2O that can be produced from 61.3 g O2 and excess NH3. For O2, the molar mass is 2(15.999 g/mol)=31.998 g/mol, and there are six mol H2O produced per five mol O2that react. 61.3 g O2×1 mol O231.998 g O2×6 mol H2O5 mol O2×18.0148 g H2O1 mol H2O=41.4 g H2O The maximum amount of product is the smaller of these two masses. O2 is the limiting reactant because it limits the amount of product that can form.

What is the maximum mass of S8 that can be produced by combining 77.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

First, calculate the amount of S8 that can be produced from 77.0 g SO2 and excess H2S. SO2 has a molar mass of 32.066+2(15.999)=64.064 g/mol. There are 3 mol S8 produced per 8 mol SO2 that react. S8has a molar mass of 8(32.066)=256.528 g/mol. 77.0 g SO2×1 mol SO264.064 g SO2×3 mol S88 mol SO2×256.528 g S81 mol S8=116 g S8 Next, calculate the amount of S8 that can be produced from 77.0 g H2S and excess SO2. H2S has a molar mass of 2(1.0079)+32.066=34.0818 g/mol. There are 3 mol S8 produced per 16 mol H2S that react. S8 has a molar mass of 8(32.066)=256.528 g/mol. 77.0 g H2S×1 mol H2S34.0818 g H2S×3 mol S816 mol H2S×256.528 g S81 mol S8=109 g S8 The amount of product that can form is the smaller of these two masses. H2S is the limiting reactant because it limits the amount of product that can form.

The chemical reaction, CaI2+2AgNO3⟶2AgI+Ca(NO3)2 what mass of silver nitrate in grams is needed to produce 5.27 mol of calcium nitrate?

First, use the given amount of calcium nitrate and the coefficients in the balanced equation to find the number of moles of silver nitrate. 5.27 mol Ca(NO3)2×2 mol AgNO31 mol Ca(NO3)2=10.5 mol AgNO3 Then, convert the moles of silver nitrate to grams of silver nitrate using its molar mass. 10.5 mol AgNO3×169.87 g AgNO31 mol=1790 g AgNO3

For the reaction 2Al+3H2SO4⟶3H2+Al2(SO4)3 how many grams of hydrogen, H2, are produced from13.7g of aluminum, Al?

First, use the molar mass of Al to find the number of moles of aluminum in 13.7 g. 13.7g Al×1 mol Al26.98 g=0.508 mol Al Then, convert to moles of hydrogen using the reaction coefficients. 0.508 mol Al×3 mol H22 mol Al=0.762 mol H2 Finally, convert moles of hydrogen to grams using the molar mass of H2. 0.762 mol H2×2.02 g H21 mol=1.54 g H2

Reducing agent

Groups 1-2 metals

Are the following salts soluble or insoluble in water? a. Na3PO4

Guideline 1: Compounds of group 1 elements (Li+, Na+, K+, Rb+, Cs+, and Fr+) and ammonium (NH4+) are soluble.

Are the following salts soluble or insoluble in water? Ag2S

Guideline 5: With the exception of guideline 1 and the barium ion (Ba2+), hydroxides (OH-) and sulfides (S2-) are insoluble.

Are the following salts soluble or insoluble in water? MgSO4

Guideline 7: With the exception of compounds of calcium (Ca2+), strontium (Sr2+), barium (Ba2+), and the ions listed in guideline 6, all sulfates are soluble. a. Guideline 1: Compounds of group 1 elements (Li+, Na+, K+, Rb+, Cs+, and Fr+) and ammonium (NH4+) are soluble.

Oxidation

Increase in oxidation state Loss of electrons Reducing agent

How many milliliters of 10.0 M HCl(aq) are needed to prepare 925.0 mL of 1.00 M HCl(aq)?

Input the initial and final concentrations of HCl along with the final volume of the diluted solution. (10.0 M)𝑉1=(1.00 M)(925.0 mL) Rearrange the equation and solve for 𝑉1. 𝑉1=(1.00 M)(925.0 mL)10.0 M=92.5 mL

Strong electrolyte

Ionic (salts) NaCl(aq), K2SO4(aq) Ionic (strong bases) NaOH(aq), KOH(aq) Strong acid HCl(aq), HNO3(aq)

Oxidizing agent

Ions of group 1-2 metals

The oxidation state of P in K3PO4

K3 P O4 3(+1) + x + 4(−2) = 0 3+ x- 8= 0 x = 8 − 3 = +5

Complete the equation for the dissociation of K3PO4(aq). Omit water from the equation because it is understood to be present.

K3PO4(aq)⟶3K+(aq)+PO3−4(aq)

the most active metal.

Li

the strongest reducing agent.

Li

What concentration of NO−3 results when 701 mL of 0.299 M NaNO3 is mixed with 919 mL of 0.345 M Ca(NO3)2?

Molarity (𝑀) is defined as the number of moles of solute, 𝑛, divided by the volume of solution in liters, 𝑉. 𝑀=𝑛𝑉 Start by determining the number of moles of ion contributed by each compound. Convert the molarity and volume of each compound to moles. Then, convert moles of each compound to moles of NO−3 using the formula unit ratio. 0.299 mol NaNO31 L×(1 L1000 mL×701 mL)×1 mol NO−31 mol NaNO3=0.210 mol NO−3 0.345 mol Ca(NO3)21 L×(1 L1000 mL×919 mL)×2 molNO−31 mol Ca(NO3)2=0.634 mol NO−3 Add these values to determine the total number of moles of NO−3. 0.210 mol NO−3+0.634 mol NO−3=0.844 mol NO−3 Next, determine the total volume of the solution, in liters. (701+919) mL×1 L1000 mL=1.62 L Finally, calculate molarity. 𝑀=𝑛𝑉=0.844 mol1.62 L=0.521 M

Non-electrolyte

Molecular (most) C6H12O6(aq) and other sugars

Write the net ionic equation for the following overall equation. Ca2+(aq) + 2 NO3-(aq) + 2 K+(aq) + SO42-(aq) →CaSO4(s) + 2 K+(aq) + 2 NO3-(aq)

Net Equation: Ca2+(aq) + SO42-(aq) → CaSO4(s)

H2NO3

Nitric acid

NO2-

Nitrite

HNO2

Nitrous acid

Write the ionic and net ionic equations: Write the ionic and net ionic equations: Total Balanced 2 NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s) + 2 NaNO3(aq) • Ionic equation2 Na+(aq) + 2 OH-(aq) + Pb2+(aq) + 2 NO3-(aq) → Pb(OH)2(s) + 2 Na+(aq) + 2 NO3-(aq)

Pb2+(aq) + 2 OH-(aq) → Pb(OH)2(s)

𝒘 = −𝑷𝚫𝑽

Pressure-volume work is work caused by a volume change against an external pressure. The relationship between work, pressure, and a change in volume is expressed as

Gaseous hydrogen and oxygen can be prepared in the laboratory from the decomposition of gaseous water. The equation for the reaction is 2H2O(g)⟶2H2(g)+O2(g) Calculate how many grams of O2(g) can be produced from 39.0 g H2O(g).

Start by converting the mass of H2O to moles. The molar mass of H2O is 2(1.0079)+15.999=18.0148 g/mol. 39.0 g H2O×1 mol H2O18.0148 g H2O=2.16 mol H2O Next, use the coefficients in the balanced chemical equation to relate the number of moles of H2O to the number of moles of O2. In this case, 1 mol O2 is produced per 2 mol H2O. 2.16 mol H2O×1 mol O22 mol H2O=1.08 mol O2 Finally, convert the number of moles of O2 to grams. The molar mass of O2 is 2(15.999)=31.998 g/mol. 1.08 mol O2×31.998 g O21 mol O2=34.6 g O2

Write the ionic and net ionic equations for the following acid-base reactions. 2 H+(aq) + 2 ClO4-(aq) + Ba2+(aq) + 2 OH-(aq) → Ba2+(aq) + 2 ClO4-(aq) + 2 H2O(l)

Strong Acid H+(aq) + OH-(aq) → H2O(l)

Identify any redox processes. a. 2 H2(g) + O2(g) → 2 H2O(g)

Synthesis (combination) redox

2 H2(g) + O2(g) → 2 H2O(l)

Synthesis reaction; oxidation- reduction is the driving force.

Combining 0.361 mol Fe2O3 with excess carbon produced 12.5 g Fe. Fe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles? What is the theoretical yield of iron in moles? What is the percent yield?

The actual yield of iron is 12.5 g. Convert this value to moles using the molar mass of iron, 55.847 g/mol. 12.5 g Fe×1 mol55.847 g=0.224 mol Fe The theoretical yield of Fe is based on the given amount of Fe2O3. If all 0.361 mol Fe2O3 reacted to give iron, then double that number of moles of Fe is produced according to the coefficients from the balanced chemical equation. 0.361 mol Fe2O3×2 mol Fe1 mol Fe2O3=0.722 mol Fe The percent yield is found by dividing the actual yield by the theoretical yield and then multiplying by 100%. percent yield=actual yieldtheoretical yield×100%=0.224 mol0.722 mol×100%=31.0% When calculating the percent yield, the units of the actual and theoretical yields do not matter, so long as they are the same (i.e., both moles or both grams).

A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 15.92 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

The balanced equation for the precipitation reaction is Pb(NO3)2(aq)+2NaCl(aq)⟶PbCl2(s)+2NaNO3(aq) Start by determining the number of moles of PbCl2(s) that precipitated. (15.92 g)(1 mol PbCl2278.1 g PbCl2)=0.05725 mol PbCl2 Convert the moles of PbCl2 to the moles of Pb(NO3)2 using the coefficients in the balanced equation. The mole ratio of Pb(NO3)2 to PbCl2 is 1:1. (0.05725 mol PbCl2)(1 mol Pb(NO3)21 mol PbCl2)=0.05725 mol Pb(NO3)2 Calculate the molarity of the solution by dividing the moles of Pb(NO3)2 by the volume of the solution in liters. 𝑀=𝑛𝑉=0.05725 mol Pb(NO3)20.2000 L=0.2862 M Pb(NO3)2

𝚫𝑯 = 𝒒𝐩

The change in enthalpy of a system, ΔH, is equal to the flow of heat at constant pressure, qp, expressed as

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1047 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 14.43 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

The equation for the titration reaction is 2NaOH(aq)+H2SO4(aq)⟶Na2SO4(aq)+2H2O(l) Use the volume of NaOH added, the concentration of the NaOH solution, and the stoichiometric coefficients from the titration reaction to determine the number of moles of H2SO4 in the diluted acid solution. Divide by the volume of the dilute acid solution to determine the concentration of the dilute acid solution. 14.43 mL NaOH×0.1047 mol NaOH1L×1mol H2SO42mol NaOH10.00 mL H2SO4=0.07554 M H2SO4 Then, determine the concentration of the original acid solution. 250.0 mL× 0.07554 M H2SO425.00 mL=0.7554 M H2SO4

If 5.13 g of CuNO3 is dissolved in water to make a 0.790 M solution, what is the volume of the solution in milliliters?

The molar mass of CuNO3 is (63.546 g/mol)+(14.007 g/mol)+3(15.999 g/mol)=125.55 g/mol The number of moles of CuNO3 is 5.13 g×1125.55 g/mol=0.0409 mol Molarity, 𝑀, is the number of moles, 𝑛, per liter of solution, 𝑉. 𝑀𝑀𝑉𝑉=𝑛𝑉=𝑛=𝑛𝑀=0.0409 mol0.790 M=0.0517 L=51.7 mL

By titration, it is found that 65.9 mL of 0.171 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.

The number of millimoles of NaOH required to neutralize the HCl(aq) solution is millimoles of NaOH=(0.171 mmol1 mL)(65.9 ml)=11.3 mmol From the equation of the neutralization reaction NaOH(aq)+HCl(aq)⟶NaCl(aq)+H2O(l) it requires 1 mol NaOH(aq) to neutralize 1 mol HCl(aq). Thus, millimoles of HCl=millimoles of NaOH=11.3 mmol The concentration of the HCl(aq) solution is 𝑀=𝑛𝑉=11.3 mmol25.0 mL=0.451 M

internal energy

The relationship between a system's internal energy change (work and heat) is expressed in the equation: ΔU = q + w.

Specific Heat - Depends on Phase

The specific heat, c, of a substance is the number of calories required to raise exactly 1 g of the substance by exactly 1°C. J/g ∙ °C

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g) If 15.4 kg of Al2O3(s), 52.4 kg of NaOH(l), and 52.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced? mass of cryolite produced: Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete?

To balance the equation, start by balancing Al, then Na and F. Save the coefficient of H2O for last. Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g) To determine the mass of cryolite produced, you must determine which reactant is the limiting reactant. The number of moles of Al2O3 is (15.4 kg)(103 g1 kg)(1 mol Al2O3101.96 g Al2O3)=151 mol Al2O3 The number of moles of NaOH is (52.4 kg)(103 g1 kg)(1 mol NaOH40.00 g NaOH)=1310 mol NaOH The number of moles of HF is (52.4 kg)(103 g1 kg)(1 mol HF20.01 g HF)=2620 mol HF Because 1 mol Al2O3 reacts with 6 mol NaOH and 12 mol HF, 151 mol Al2O3 requires 6(151 mol)=906 mol NaOHand 12(151 mol)=1810 mol HF. Because there are more moles of NaOH and HF available than this, Al2O3 is the limiting reactant. The mass of cryolite produced is (151 mol Al2O3)(2 mol Na3AlF61 mol Al2O3)(209.95 g Na3AlF61 mol Na3AlF6)(1 kg103 g)=63.4 kg Na3AlF6 To determine the amount of excess reactants, determine the amount of each reactant that is used and subtract it from the initial amount of reactant. (52.4 kg NaOH)−(906 mol NaOH×40.00 g1 mol×1 kg1000 g)=16.2 kg NaOH (52.4 kg HF)−(1810 mol HF×20.01 g1 mol×1 kg1000 g)=16.1 kg HF In total, the excess is (16.2+16.1)kg=32.3 kg

Writing Ionic and Net Ionic Equations for Acid-Base Reactions

Total balanced equation; Ionic equation; Net ionic equation • The method is similar to that for precipitation reactions. • Write all strong electrolytes as separate ions. • The net ionic equation shows only those components that have changed during the reaction.

Reaction of Weak Acid with Strong Base HF(aq) + K+(aq) + OH-(aq) → K+(aq) + F-(aq) + H2O(l)

Weak Acid HF(aq) + OH-(aq) → F-(aq) + H2O(l)

Write the ionic and net ionic equations for the following acid- base reactions. HC2H3O2(aq) + Na+(aq) + OH-(aq) → Na+(aq) + C2H3O2-(aq) + H2O(l)

Weak Acid HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l)

Weak electrolyte

Weak acid HNO2(aq), H3PO4(aq) Weak base NH3(aq), CH3NH2(aq)

Write the equations for the pairs that will react. Li(s) + Cu2+(aq) → 2 Li+(aq) + Cu(s)

Will react

Consider the combustion reaction for octane (C8H18), which is a primary component of gasoline. 2C8H18+25O2⟶16CO2+18H2O How many moles of CO2 are emitted into the atmosphere when 26.6 g C8H18 is burned?

You are given grams of the reactant, C8H18. You will need to convert that into moles of C8H18 using its molar mass. 26.6 g C8H18×1 mol C8H18114.23 g C8H18=0.233 mol C8H18 Now, consider the molar ratio between C8H18 and CO2. According to the balanced chemical equation, 2 mol C8H18 is required to produce 16 mol CO2. Use this ratio to convert from moles of C8H18 to moles of CO2. 0.233 mol C8H18×16 mol CO22 mol C8H18=1.86 mol CO2

Determine the oxidation state of each of the following species. a. Mg b. P3- c. Fe in FeCl3

a. 0 (rule 1) A neutral element that is not part of a compound has an oxidation state of zero. b. −3 (rule 2) Monoatomic ions have oxidation states equal to their ionic charges. c. +3 (rule 2) Monoatomic ions have oxidation states equal to their ionic charges.

Classify each of these reactions: Ba(OH)2(aq)+2HNO2(aq)⟶Ba(NO2)2(aq)+2H2O(l)

acid-base neutralization

Common states

are solid (s), liquid (l), gas(g), and aqueous (aq).

reactions involve the transfer of electrons.

are used to keep track of the electron transfer in redox reactions.

Write the balanced equation for the neutralization reaction between H3PO4 and NaOH in an aqueous solution. Phases are optional.

balanced equation: H3PO4+3NaOH⟶Na3PO4+3H2O

Hess's law

calculate the enthalpies of different reactions

Energy...

can take the form of heat or work and can be transferred between a system and the surroundings.

CO3-2

carbonate

ClO3-

chlorate

HClO3

chloric acid

ClO2-

chlorite

HClO2

chlorous acid

Classify each of the reactions: CaO+H2O⟶Ca(OH)2

combination

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)

combustion

CaCO3(s) → CaO(s) + CO2(g)

decomposition

Classify each of the reactions: C12H22O11⟶12C+11H2O

decomposition

percent yield

defined as 100% times the ratio of the actual yield to the theoretical yield.

Work..

done on the system is mathematically defined as positive in value, whereas work done by the system is defined as negative in value.

Classify each of the reactions: Pb(NO3)2+NiCl2⟶PbCl2+Ni(NO3)2

double replacement

MgI2(aq) + 2 AgNO3(aq) → 2 AgI(s) + Mg(NO3)2(aq)

double-replacement

Aqueous solutions of silver nitrate and sodium phosphate are mixed together, forming solid silver phosphate and aqueous sodium nitrate. Write the balanced chemical equation for the reaction. Phases are optional.

equation: 3AgNO3+Na3PO4⟶3NaNO3+Ag3PO4

Write and balance the equation for the complete combustion of heptane, C7H16. Phase symbols and energy changes are optional.

equation: C7H16+11O2⟶7CO2+8H2O

Solid iron(III) oxide reacts with hydrogen gas to form solid iron and liquid water. Write the balanced chemical equation for the reaction described. Phases are optional.

equation: Fe2O3+3H2⟶2Fe+3H2O

Aqueous sodium sulfate reacts with aqueous barium bromide to form aqueous sodium bromide and solid barium sulfate. Write the balanced chemical equation for the reaction. Include physical states.

equation: Na2SO4(aq)+BaBr2(aq)⟶2NaBr(aq)+BaSO4(s)

Chlorides (Cl-), bromides (Br-), and iodides (I-) are soluble,

except for those of Ag+, Pb2+, and Hg22+.

ClO-

hypochlorite

HClO2

hypochlorous acid

𝒘 = 𝟎

inside a rigid container of the bomb (constant volume)

With the exception of guideline 2, silver (Ag+), mercury (Hg22+), and lead (Pb2+) salts are

insoluble

Except for compounds of the cations in guideline 1, carbonates (CO32-), sulfites (SO32-), phosphates (PO43-), and chromates (CrO42-) are

insoluble.

With the exception of guideline 1 and the barium ion (Ba2+), hydroxides (OH-) and sulfides (S2-) are

insoluble.

U =

internal energy of a system defined as the sum of all kinetic and potential energies.

𝐌 𝐕 = 𝐌 𝐕

is an equation that can be used for dilutions

Enthalpy Hrxn

is generally measured by calorimetry using 𝑞 = 𝑚𝑐∆𝑇 reported in units of kJ/mol (reactant used, or product produced)

The enthalpy of solution

is the amount of heat absorbed or given off during the process.

actual yield

is the amount of product that is obtained in an experiment

Stoichiometry

is the calculation of quantities of any substances involved in a chemical reaction from the quantities of the other substances via the mole

ΔHrxn

is the enthalpy change for a specific chemical reaction occurring in stoichiometric quantities.

ΔHfus

is the enthalpy change for the process of fusion. (melting)

ΔHvap

is the enthalpy change for the process of vaporization

ΔHsol

is the enthalpy change when a solid dissolves in a solvent

theoretical yield

is the maximum amount of product that can be formed

Calorimetry

is the study of heat transfers by measuring the temperature changes in the substances involved.

MgCO3 (s) heat -> MgO(s) +CO2 (g)

magnesium carbonate to decompose to magnesium oxide and carbon dioxide.

Write the balanced molecular equation and net ionic equation for the neutralization reaction between hydrochloric acid and strontium hydroxide. Include the phase of each species. molecular equation: net ionic equation:

molecular equation: 2HCl(aq)+Sr(OH)2(aq)⟶2H2O(l)+SrCl2(aq) net ionic equation: H+(aq)+OH−(aq)⟶H2O(l)

Complete and balance the molecular equation for the reaction of aqueous chromium(II) bromide, CrBr2, and aqueous sodium carbonate, Na2CO3. Include physical states. molecular equation: Enter the balanced net ionic equation for the reaction. Include physical states. net ionic equation:

molecular equation: CrBr2(aq)+Na2CO3(aq)⟶2NaBr(aq)+CrCO3(s) net ionic equation: Cr2+(aq)+CO2−3(aq)⟶CrCO3(s)

Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2. Include physical states. molecular equation: Enter the balanced net ionic equation for this reaction. Include physical states. net ionic equation:

molecular equation: Na2SO4(aq)+Ba(NO3)2(aq)⟶2NaNO3(aq)+BaSO4(s) net ionic equation: Ba2+(aq)+SO2−4(aq)⟶BaSO4(s)

How many moles of NaOH are present in 18.0 mL of 0.150 M NaOH?

moles=molarity× liters Convert the volume from milliliters to liters. 18.0 mL×1 L1000 mL=0.0180 L Use the values of molarity and volume to calculate the number of moles of NaOH. moles=molarity× liters=(0.150 M)(0.0180 L)=0.00270 mol

Equations (equal)

must balanced based on chemical symbols.

For the chemical reaction HClO4(aq)+NaOH(aq)⟶H2O(l)+NaClO4(aq) write the net ionic equation, including the phases. net ionic equation: Which ions are considered spectator ions for this reaction?

net ionic equation: H+(aq)+OH−(aq)⟶H2O(l) Na+ ClO−4

Acid - base

neutralization

NO3-

nitrate

Write the equations for the pairs that will react. Cu(s) + Li+(aq)

no reaction (Li is more active than Cu)

Classify each of these reactions: 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq)

none of the above

Classify each of these reactions: CaO(s)+CO2(g)⟶CaCO3(s)

none of the above

Classify the following water-soluble substances: C6H12O6

nonelectrolyte

ClO4-

perchlorate

HClO4

perchloric acid

PO4-3

phosphate

PO3-3

phosphite

H3PO4

phosphoric acid

H3PO3

phosphorous acid

Parentheses

physical states of the reactants and products.

Classify each of these reactions: KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s)

precipitation

Double-Replacement

precipitation

Expressions and Meanings of 𝒒

q + The system gains heat. q - The system loses heat.

Oxidation-reduction (redox)

reactions involve the transfer of electrons.

Reactants

rearrange their bonding to form products.

Classify each of these reactions: Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g)

redox

Decomposition

redox

Single Replacement

redox

Synthesis (Combination)

redox

Classify each of the reactions: ZnSO4+Mg⟶Zn+MgSO4

single replacement

With the exception of compounds of calcium (Ca2+), strontium (Sr2+), barium (Ba2+), and the ions listed in guideline 6, all sulfates are

soluble

Compounds of group 1 elements (Li+, Na+, K+, Rb+, Cs+, and Fr+) and ammonium (NH4+) are

soluble.

Nitrates (NO3-), chlorates (ClO3-), perchlorates (ClO4-), and acetates (C2H3O2-) are

soluble.

Classify the following water-soluble substances: CaCl2

strong electrolyte

Classify the following water-soluble substances: HBr

strong electrolyte

Classify the following water-soluble substances: KOH

strong electrolyte

H2SO4

sulfuric acid

H2SO3

sulfurous acid

SO4-2

sulphate

SO3-2

sulphite

2 H2(g) + O2(g) → 2 H2O(l)

synthesis

Coefficients

tell you in what proportions the reaction occurs.

Chemical formulas

tell you what happens during the reaction.

Coinage metals are..

the among the least active (least easily oxidized).

Changing the coefficients changes

the amounts of the compounds.

Changing the subscripts changes

the identities of the compounds.

m represents

the number of moles of each product

n represents

the number of moles of each reactant

Heat..

transferred into the system is mathematically defined as positive in value, whereas heat transferred from the system to the surroundings is defined as negative in value.

Expressions and Meanings of 𝒘

w + Work is done on the system. w - Work is done by the system.

Classify the following water-soluble substances: HNO2

weak electrolyte

Classify the following water-soluble substances: NH3 (weak base)

weak electrolyte

𝚫𝑼 = 𝒒𝐯

which is the heat flow at constant volume

Expressions and Meanings of ∆𝑼

ΔU + The system gains internal energy. ΔU - The system loses internal energy.

Solutions and Concentration

• A solution is a homogeneous mixture of two or more substances. • The solvent dissolves substances called solutes. • Concentration expresses the quantity of a solute in its solvent. • Solutions with a large amount of solute per volume of solvent are concentrated. • Solutions with a small amount of solute per volume of solvent are dilute.

state function

• A state function describes the current state of a system and is independent of the path taken to achieve its value. • Internal energy is a state function.

Reactions of Metals with Acids

• Acids produce H+ in aqueous solution. • H appears in the activity series • Neutral metals that appear above H in the activity series can transfer electrons to H+(aq) to form H2(g). • Neutral metals below H in the activity series do not react with acids.

Electrolytes

• Aqueous solutions of ionic compounds can conduct electricity due to the mobile hydrated ions. • Electrolytes are substances that, when dissolved in water, conduct electricity. • Ionic compounds are known as strong electrolytes because they dissociate 100% to produce solutions that conduct electricity readily.

Bomb Calorimetry Calculations

• Calculations of bomb calorimetry use the total heat capacity of the calorimeter components and the water, Ccal, not just the specific heat of water, c . water • Ccal is measured very accurately for each calorimeter for a given volume of water. Once Ccal is known for a particular unit, measurements can made using regarding these relationships:

Combustion Reactions

• Combustion is the rapid combination of a substance with oxygen. • Hydrocarbon fuels undergo combustion to produce carbon dioxide and water. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

Precipitation Reactions

• Compound are either soluble or insoluble in water.• Precipitation = driving force for double replacement reactions. MgI2(aq) + 2 AgNO3(aq) → 2 AgI(s) + Mg(NO3)2(aq)

Constant-Pressure Calorimetry

• Constant-pressure calorimetry is commonly carried out in undergraduate laboratories using Styrofoam© cups and a lid. • This minimizes heat transfer between the system and surroundings.• The system is not sealed, so the calorimeter pressure is the atmospheric pressure. • This type of system makes it easy to measure qp for any heat transfer that can happen in aqueous solution. • For chemical reactions, 𝒒𝐩 = ∆𝐻, the enthalpy change for the reaction.

Constant-Volume Calorimetry

• Constant-volume calorimetry is carried out in a bomb calorimeter. • They are designed to withstand high temperature and pressure changes and to be sealed and insulated from the surroundings. • Bomb calorimeters are close to a true isolated system.

Decomposition Reactions

• Decomposition occurs when a reactant breaks down to less complex products. • Water decomposes, by means of electrical energy, into H2 gas and O2 gas. 2 H2O(l) → 2 H2(g) + O2(g) KClO3(s) → KCl(s) + O2(g)

Dilution

• Dilution is the process of adding more solvent to decrease the concentration of solute in a solution.

Driving Forces and Reaction Types

• Double-replacement reactions can form solids (called precipitates), which reduce energy. • Neutralization is the driving force for acid-base reactions that produce a salt and, often, water. • Single-replacement, synthesis, and decomposition reactions transfer electrons to form lower-energy products. This driving force is known as oxidation-reduction.

Enthalpy Change, ∆𝑯

• Enthalpy change, ΔH, is equal to the heat transfer for a process carried out at constant pressure. 𝒒𝐩 = ∆𝐻 • ΔH is a state function and therefore independent of the path taken from the initial to final state. • ΔH depends only on what the reactants and products are, and not on the chemical reactions carried out to get from one to the other.

Energy and Enthalpy

• Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume. • Pressure-volume work is work done when there is a volume change as measured against an external pressure. • A constant pressure system allows the piston to move without resistance.

is the amount of heat absorbed or given off during the process.

• Heat changes occur during solution formation, which can be endothermic or exothermic, depending on the solute. • The enthalpy of solution is the amount of heat absorbed or given off during the process.

Mole Calculations for Chemical Reactions

• In chapter 3 the mole was used to calculate mass and the number of units based on Avogadro's number for chemical substances. • In chapter 5 the mole will be applied to chemical transformations where stoichiometry refers to the ratios of the substances in chemical reactions. (Coefficients of balanced reactions) • Stoichiometry is the calculation of quantities of any substances involved in a chemical reaction from the quantities of the other substances.

Sodium Chloride in Water

• Ionic compounds dissociate into individual ions and disperse among the water molecules. -NaCl forms separate Na+ and Cl- ions surrounded by water molecules, and are known as hydrated ions. -Using (aq) after the formula of an ionic compound indicates that the compound is present as hydrated ions. -NaCl(s) H2O NaCl(aq) → Na+(aq) + Cl-(aq)

Ionic Compounds in Water

• Many reactions only occur when in aqueous solution. As solids they do not react. • Dissolved compounds are more mobile than when in their solid state and, therefore, more likely to come into contact. • Ionic compound dissociate to form independent ions in solution. NaCl(s) H2O NaCl(aq) → Na+(aq) + Cl-(aq)

Molarity

• Molarity is defined as the number of moles of solute per liter of solution: molarity = moles of solute liter of solution • The unit of molarity is molar, symbolized as M. • 0.50 M NaCl means that there are 0.50 moles of NaCl in each 1.0 L of the solution.

Enthalpy

• Most chemical reactions are performed in open systems with exchange of both heat and work between system and surroundings. • We are interested in the flow of heat in systems at constant pressure. (atmospheric) • Enthalpy (H) is a state function that relates internal energy, pressure, and volume.

Molecular Compounds in Water

• Most molecular compounds that dissolve in water form solutions that do not conduct electricity. They are called nonelectrolytes. • Nonelectrolytes dissolve as molecules (hydrated).• Acids are molecules that ionize when dissolved in water to produce solutions that conduct electricity.HCl(aq) → H+(aq) + Cl-(aq) • Strong acids ionize 100% and are strong electrolytes.

Reaction of NaCl(aq) and AgNO3(aq)

• NaCl and AgNO3 ionize in water, forming solutions of hydrated ions. AgNO3(aq) + NaCl(aq) → Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) • Any oppositely charged ions will attract each other. Any that, together, form an insoluble substance will precipitate. Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + NaNO3(aq)

Reaction of Strong Acid with Strong Base H+(aq) + Cl-(aq) + K+(aq) + OH-(aq) → K+(aq) + Cl-(aq) + H2O(l)

• Net ionic equation (same for all strong acid-base reactions) H+(aq) + OH-(aq) → H2O(l)

Driving Forces for Reactions in Aqueous Solutions

• Reactions occur spontaneously due to a combination of changes in heat energy (enthalpy) and randomness (entropy), which are discussed in future chapters. • Spontaneous reactions follow typical patterns classified by their driving force. • Driving forces are associated with the formation of stable, lower- energy products.

Using Oxidation States to Identify Redox Reactions

• Redox reactions involve the transfer of electrons.• One reactant loses electrons (oxidation), while another gains electrons (reduction).• Oxidation occurs when the oxidation state of an element increases. • Reduction occurs when the oxidation state decreases.

Ways to Identify a Limiting Reactant

• Separately calculate the amount of product that could form from each reactant. The reactant that produces the smaller amount of product is limiting. • Divide the moles of each reactant by its coefficient. The reactant with the smallest quotient is the limiting reactant.

Solubility Guidelines

• Solubility is determined empirically.• Collect solubility data in a chart.

Interpreting Specific Heat Values

• Substances with higher specific heat values resist temperature changes more than substances with lower specific heat values. • Examples:• Due to its high specific heat of 4.184 J/g · °C, water can absorb a significant amount of heat without experiencing an increase in temperature. • Dry sand (the major component of sand, silicon dioxide, has a specific heat of 1.0 J/g · °C) on a beach can be uncomfortably hot to walk on barefoot during a summer afternoon, while the water is cool.

Synthesis Reactions

• Synthetic reactions involve combining reactants to form a more complex product. • Example: Iron metal reacts with oxygen to form iron(III) oxide. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

Limiting Reactant Terminology

• The reactant that is used up is the limiting reactant and is present in limiting quantity. • Any reactant that is left over is present in excess. • The reaction has gone to completion when the limiting reactant is used up. • When doing calculations use the limiting reactant.

path functions

• The value of path functions depends on the path, or the sequence of steps taken between initial and final states. • Work and heat are path functions.

Acid-Base Reactions

• These reactions are a type of double-replacement reaction. -Usually, no visible change occurs in the reaction mixture. -Heat is released. -The hydrogen from the acid combines with hydroxide from the base to form water, while the remaining ions form a salt.

Single-Replacement Reactions

• This reaction, also known asa displacement reaction, occurs when an element replaces one of the elements in that compound. • In the reaction of zinc metal with hydrochloric acid, Zn replaces H in the compound.

Calorimetry - Heat Transfer

• When a piece of hot metal is placed in cold water, the metal cools and the water warms. System (metal) Surrounding (water) • The amount of heat lost by the metal is equal to the heat gained by the water (assuming no heat transfer with the outside surroundings). 𝒒 = 𝒎𝒄𝜟𝑻 𝐪 = 𝟎 = (𝐦water)(𝐜𝑤𝑎𝑡𝑒𝑟)(𝚫𝐓𝑤𝑎𝑡𝑒𝑟) + (𝐦metal)(𝐜metal)(𝚫𝐓metal) 𝒒𝒔𝒐𝒍𝒏 = −𝒒𝒎

A Dilution Equation

• When a solution is diluted with solvent, the number of moles of solute does not change. • Before the dilution, M1 × V1 = mol of solute. • After the dilution, M2 × V2 = mol of solute. • Because the number of moles of solute has not changed, these two expressions are equal to one another: 𝐌𝐕=𝐌𝐕 𝟏𝟏 𝟐𝟐

Energy, Heat, and Work

• Work, w, is the energy resulting from a force acting on an object over a distance. w = f x d (mechanical) • The flow of energy that causes a temperature change in an object or its surroundings is known as heat, q. w ~ q (chemical) • Both work and heat can be exchanged between the system and surroundings. (chemical transformation)

Strong Bases

• are ionic compounds containing hydroxide ions. • dissociate 100% in water • are strong electrolytes.

Weak Bases

• are molecular compounds, such as ammonia. • react with water to a small extent to produce hydroxide ions in water. • are weak electrolytes

The temperature change depends upon three things

• the identity of the solute, • the amount of solute, and • the amount of water.

exothermic processes

• Δ𝐻 < 0 when heat is transferred from the system to the surroundings. • Reactions that transfer heat to the surroundings are known as exothermic processes.

endothermic processes

• Δ𝐻 > 0 when heat is transferred from the surroundings to the system. • Reactions that absorb heat from the surroundings are known as endothermic processes.

When the reaction involves multiple moles of products and reactants

∆𝐻∘ = Ʃ𝒎[∆H(products)]−Ʃ𝒏[∆H(reactants)]

𝒒𝒗

−𝑪𝒄𝒂𝒍𝜟𝑻

What is the concentration of a solution made by diluting 35 mL of 6.0 M HCl to a final volume of 750 mL?

𝑀2=𝑀1𝑉1𝑉2=(6.0 M)(35 mL)750 mL=0.28 M

Calculate the specific heat c of water if 83.68 J is required to raise the temperature of 4.000 g of water by 5.000°C.

𝒒 = 𝒎𝒄𝜟𝑻83.68 J = (4.000 g)(𝒄)(5.000°C) 𝒄= 83.68J =4.184 J 4.000 g 5.000°C g ⋅ °C

Surroundings ->

𝒒𝐬𝐨𝐥𝐧 = 𝑪𝐜𝐚𝐥𝚫𝑻

System ->

𝚫𝑼𝐫𝐱𝐧 = 𝒒𝐯