Chemistry 131 Exam 2 Review
How many He atoms are in 0.568 moles of He atoms?
(0.568 times 6.02x10^23)= 3.42x10^23 He ions
CH4 + 2 H2O → 4 H2 + CO2 How many grams of CH4 are consumed in a process where 1.43 g H2 is produced?
(1.43 g H2)= (1 mol H2)/(2.02 g H2) (1 mol CH4)/(4 mol H2) (16.05 g CH4)/(1 mol CH4) = 2.84 g CH4
How many moles of C10H22 are in 40.1 g of C10H22?
(10)(12.01)+(22)(1.008) = 142.3 g/mol (40.1 divided by 142.3)= 2.28x10^-1 mol C10H22
How many C16H30 molecules are in 70.3 g of C16H30?
(16)(12.01)+(30)(1.008) = 222.4 g/mol (70.3 divided by 222.4)=0.3161 mol C16H30
The molar mass of MgSO4 is
(1x24.31 + 1x32.07 + 4x16.00)=120.4 g/mol
How many moles of O2 molecules are in 3.11x1023 O2 molecules?
(3.11x10^23 divided by 6.02x10^23)=0.516 mol O2 molecules
How many moles of He atoms are in 4.76x1023 He atoms?
(4.76x10^23 divided by 6.02x10^23)=0.790 mol He atoms
CH4 + 2 H2O → 4 H2 + CO2 How many grams of H2 is produced from 5.34 g H2O ?
(5.34 g H2O)= (1 mol H2O)/(18.02 g H2O) (4 mol H2)/(2 mol H2O) (2.02 g H2)/(1 mol H2) = 1.20 g H2
Which of the following is a (conjugate base, conjugate acid) pair?
(CN-, HCN) ***The species formed when a proton is removed from an acid is referred to as the conjugate base of that acid, i.e. B- is the conjugate base of HB. The species formed when a proton is added to a base is called the conjugate acid of that base, i.e. HA is the conjugate acid of A-.
Which of the following is a (conjugate acid, conjugate base) pair?
(H2O, OH-)
What is the oxidation number of Fe in iron(II) chloride?
+2 ***Example: If simply asked for the oxidation number of iron or Fe, the answer is zero. If asked for the oxidation number of Fe in a compound, the answer is equal to its charge. When an atom of Fe becomes part of a compound it always becomes a positive ion (Fe2+ or Fe3+; you need to figure this out from the name or formula of the compound)
What is the oxidation number of Mg in MgO?
+2 ***the oxidation number of a monatomic ion is equal to its charge. Example: If simply asked for the oxidation number of sodium or Na, the answer is zero. If asked for the oxidation number of sodium ion, the answer is +1. If asked for the oxidation number of Na in a compound, the answer is +1. When a sodium atom becomes part of a compound it always becomes Na+
What is the oxidation number of Cr in Cr(NO3)3?
+3
What is the oxidation number of C in CF4?
+4 F is more electronegative; its oxidation number is -1 THEN x+4(-1)=0. Solve for x.
What is the oxidation number of S in sodium sulfite?
+4 Sulfite is SO32-. Therefore: x+(3)(-2)=-2; solve for x
What is the oxidation number of Cl in perchlorate?
+7 Perchlorate is ClO4-. Therefore: x+(4)(-2)=-1; solve for x. ***In oxyanions, the oxidation number of (each) oxygen is -2. The total oxidation number for all the atoms in a polyatomic ion must add up to the charge of the ion. In the equation above, we let x=oxidation number of the central atom.
The amount of acetic acid (HC2H3O2) in vinegar can be determined by titrating it with NaOH solution. The net ionic equation is HC2H3O2(aq) + OH-(aq) → H2O(l) + C2H3O2-(aq) Calculate the molarity of acetic acid in a 20.00-mL sample of vinegar if 40.31 mL of 0.3729 M NaOH is required to reach the equivalence point.
.751 M millimoles NaOH used up = (Molarity of NaOH solution) times (Volume of NaOH solution) = (0.3729 mol/L) (40.31 mL) = 15.03 mmol The net ionic equation tells us that the mole-to-mole ratio of acetic acid (HC2H3O2) to NaOH is 1:1. Therefore, the amount of acetic acid in the sample is also 15.03 mmol. The molarity of HC2H3O2 in the vinegar sample is obtained by dividing moles of HC2H3O2 by the volume of vinegar in liters; or dividing millimoles of HC2H3O2 by the volume of vinegar in milliliters. M= (15.03/20.00)=).751
A 1.55-L sample of aqueous solution contains 0.643 g KCl and 0.378 g KNO3. What is the molarity of the potassium ions in the solution?
0.00789 M First we need to find moles of K+ ions. In one mole of KCl, we have 1 mol K+. In one mole of KNO3, we have 1 mol K+. Therefore: n = 1 times (0.643/74.55) + 1 times (0.378/101.1) mol K+ Note the numbers in parentheses above are the moles of KCl and KNO3, respectively. Dividing grams by molar mass gives us moles.
What is the molar concentration of nitrate ions in a 0.205M sodium nitrate(aq) solution?
0.205 You need to know the formula of sodium nitrate; the formula tells you the number of moles of nitrate ions there are in one mole of sodium nitrate. It's 1. Therefore, the molar concentration of nitrate ions is just 1 TIMES the molar concentration of sodium nitrate. 1 times 0.205 equals 0.205.
What is the molar concentration of sodium ions in a 0.406M sodium chloride(aq) solution?
0.406 mol Na+/L You need to know the formula of sodium chloride; the formula tells you the number of moles of sodium ions there are in one mole of sodium chloride. It's 1. Therefore, the molar concentration of sodium ions is just 1 TIMES the molar concentration of sodium chloride. 1 times 0.406 equals 0.406.
The amount of acetic acid (HC2H3O2) in vinegar can be determined by titrating it with NaOH solution. The net ionic equation is HC2H3O2 + OH-(aq) → H2O(l) + C2H3O2-(aq) Calculate the amount of acetic acid (in grams) present in a sample if 41.60 mL of 0.210M NaOH is required to reach the equivalence point.
0.524 g moles NaOH used up = (Molarity of NaOH solution) times (Volume of NaOH solution) = (0.210 mol/L) (41.60 mL) = 8.736 mmol or 8.736x10-3 mol The net ionic equation tells us that the mole-to-mole ratio of acetic acid (HC2H3O2) to NaOH is 1:1. Therefore, the amount of acetic acid in the sample is also 8.736x10-3 mol. Finally, change moles to grams (molar mass of HC2H3O2 is 60.05 g/mol): 8.736x10-3 mol HC2H3O2 0.524 g HC2H3O2
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in SO3 SO2 we can balance H and O by adding H2O and/or OH- if the reaction is occurring in basic solution. When H and O are balanced in the example above, what would be the coefficient of H2O?
1
N2 + 3 H2 → 2 NH3 If 0.159 mol of N2 and 0.445 mol of H2 are mixed and the reaction were to go to completion,
1. Figure out what x is if one reactant is consumed (in this case, let's do H2 ): x=(0.445 mol/3)=0.148 mol 2. Figure out how much of the other reactant would be consumed as well (in this case, N2): x=0.148 mol 3. Compare the amount you calculated in step 2 with the original amount of the other reactant: we have more than enough (0.159 > 0.148) 4. Finally, subtract the amount used up from the original amount (0.5159-0.148)=0.011 mol
N2 + 3 H2 → 2 NH3 If 0.710 mol of N2 and 2.27 mol of H2 are mixed,
1. Figure out what x is if one reactant is consumed (in this case, let's do N2): x=0.710 mol 2. Figure out what x is if the other reactant is consumed (in this case, we do H2 ): x=(2.27/3)=0.757 mol 3. Use the smaller x value to calculate the theoretical yield. 2x=2(0.710) mol=1.42 mol
Water is added to 45.8-mL sample of 0.193 M K2SO4(aq). If the final volume is 800.0 mL, what is the molarity of the resulting solution?
1.10x10^-2 M what is the molarity of a solution prepared by diluting 45.8-mL of 0.193M K2SO4(aq) to 800.0 mL? The important thing to remember about dilution is that the amount of solute does not change -- we're just adding more solvent (water). We also expect the final concenttration to be lower. Recall that molarity is defined as M = n/V, where n=moles of solute and V=volume of solution. We can calculate moles of solute before dilution: n = MV = (0.193 mol K2SO4/L)(45.8 mL) = 8.839 mmol NOTE: since we used mL, MV above gives us millimoles (mmol). Now we calculate the molarity of the final solution. M = n/V = n/V = 8.839 mmol K2SO4 divided by 800.0 mL = 0.0110 mmol K2SO4/mL = 0.0110 mol K2SO4/L or 0.0110 M K2SO4 As expected the diluted solution has a lower concentration than the stock solution. (0.0110 < 0.193). Dilution problems can be solved using the formula M1V1 = M2V2, which essentially says that the amount of solute before dilution EQUALS amount of solute after dilution
A solution is prepared by mixing 54.6 mL of 0.790 M HCl(aq) and 25.1 mL of 0.506 M HCl(aq), then adding enough water to bring the total volume to 400.0 mL. Calculate the molarity of HCl in the resulting solution.
1.40x10^-1 M Recall that M = n/V; so to calculate moles or millimoles, we multiply M and V: millimoles HCl from first solution: (0.790 M)(54.6 mL) = 43.13 mmol millimoles HCl from second solution: (0.506 M)(25.1 mL) = 12.70 mmol Molarity of HCl = (43.13 + 12.70) mmol divided by 400.0 mL =1.40x10-1 M
A hydrocarbon is a compound containing only C and H. A pure sample of a hydrocarbon was burned in excess oxygen and 117.9 g of carbon dioxide and 32.17 g of water were collected. Which of the following is the empirical formula of the compound?
117.9 g CO2: 2.678 mol CO2 32.17 g H2O: 1.785 mol H2O One C atom in CO2, but two H atoms in H2O; C: 2.678 mol CO2 x (1 mol C/ 1 mol CO2)=2.678 mol C H: 1.785 mol H2O x (2 mol H/ 1 mol H2O)= 3.571 mol H Divide both by the smaller one C: (2.678/2.678)=1.000 H: (3.571/2.678)=1.333 Since 1.333 is not a whole #, we multiply both by 3 C: (1x3)=3 H: (1.333x3)=4 Empirical formula: C3H4
What volume of 3.69 M sodium hydroxide(aq) solution contains 17.4 grams of sodium hydroxide?
118 ml Since we are given the mass of the solute (sodium hydroxide), we must first convert it to moles: 17.4 g sodium hydroxide 0.436 mol sodium hydroxide (Review lessons on molar mass if you don't know how to do this; the molar mass of sodium hydroxide is 40.00 g/mol) Therefore: V=(.436/3.69)=0.118 L Finally we convert the volume of the solution to mL: 0.118 L 118 mL
A hydrocarbon is a compound containing only C and H. A 5.00 g pure sample of a hydrocarbon was burned in excess oxygen and 15.14 g of carbon dioxide was collected. Which of the following is the empirical formula of the compound?
15.14 g CO2: 0.3441 mol CO2 0.3441 mol C: 4.1329 g C Mass H = Total - Mass C = 5.0000 g - 4.1329 g = 0.8671 g H Divide both by the smaller one C: (0.3441/0.3441)=1.000 H: (0.8602/0.3441)=2.500 Multiply both by 2 C: (1x2)=2 H: (2.500x2)=5 Empirical formula:C2H5
How much of 0.217 M K2SO4(aq) is required in order to prepare 500.0 mL of 0.00707 M K2SO4(aq)?
16.3 mL The important thing to remember about dilution is that the amount of solute does not change -- we're just adding more solvent (water). A convenient formula to use is: M1V1 = M2V2, which essentially says that the moles of solute before dilution is equal to the moles of solute after dilution. We are asked for the volume before dilution. Therefore: Let x = volume before dilution THEN (0.217 M)(x) = (0.00707 M)(500.0 mL) Solving for x, we get: x = 16.3 mL
A pure sample of a compound is found to contain 2.09x1022 nitrogen atoms of and 4.18x1022 oxygen atoms. What is the empirical formula of the compound?
1:2 ratio, having twice as many O atoms as there are N atoms Empirical formula; NO2
How many moles of Cl atoms are in one mole of SOCl2?
2
Consider a half reaction where all the atoms are already balanced, as in 8OH- + B2Cl4 → 2BO2- + 4Cl- + 4H2O When this half reaction is finally balanced, what would be the coefficient of e-?
2 ***Example: H2O + NO3- NO2- + 2 OH- Step 1. Total charges Left side: 0 for water, -1 for nitrate ion; total charge=-1 Right side: -1 for nitrite ion; -1 for hydroxide TIMES 2; total charge =-3 Step 2. Identify side with algebraically higher charge Left side is higher: -1 is algebraically higher than -3 We need to add electrons to left side Step 3. Determine number of electrons to add High side charge MINUS Low side charge = (-1) - (-3) = 2 So, we need to add 2e- The complete balanced half reaction is 2e- + H2O + NO3- NO2- + 2 OH-
balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in Pb2+ → PbO2 we can balance H and O by adding H2O and/or H+ if the reaction is occurring in an acidic solution. What would be the coefficient of H2O when the half reaction above is balanced?
2 ***Example: NO3- → NO Step 1. The right side needs TWO more O. Add TWO H2O to that side. Now we have NO3- → NO + 2 H2O Step 2. Now the left side needs FOUR hydrogens. Add FOUR H+ to the left side. Now we have 4 H+ + NO3- → NO + 2 H2O
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in H2O → H2 we can balance H and O by adding H2O and/or OH- if the reaction is occurring in basic solution. When H and O are balanced in the example above, what would be the coefficient of H2O?
2 ***Example: NO3- → NO2- Step 1. The right side needs ONE more O. Add ONE H2O to that side. Now we have NO3- → NO2- + H2O Step 2. Now you have TWO excess hydrogens on the right. So we add TWO OH- to the right side AND TWO H2O to the left side. Now we have 2 H2O + NO3- → NO2- + H2O + 2 OH- Step 3. Examine both sides: you have 2 H2O on the left and one on the right. Simplify by subtracting (taking out) one H2O from each side. So, now we have: H2O + NO3- → NO2- + 2 OH-
When balancing a half reaction, all atoms other than H or O should be balanced first. If we were to do this for IO3- → I2 the smallest set of whole number coefficients would be
2 and 1, respectively ***You should remember that a coefficient of 1 is never written. If a formula in a chemical equation has no coefficient, it is implied to be 1
A 0.879-L sample of aqueous solution contains 0.436 moles of potassium sulfate and 1.22 moles of potassium carbonate. What is the molarity of the potassium ions in the solution?
3.77 M First we need to find moles of K+ ions. The formula of potassium sulfate tells us that in one mole of potassium sulfate, we have 2 mol K+. The formula of potassium carbonate tells us that in one mole of potassium carbonate, we have 2 mol K+. Therefore: n = (2 times 0.436 + 2 times 1.22) mol K+ then divide by 0.879
A sample of aqueous solution contains 0.952 M K2SO4 and 0.942 M K2CO3. What is the molarity of the potassium ions in the solution?
3.79 M The solution has K+ ions coming from both components. We need to calculate the concentration of K+ ions coming from each component of the solution and total the values. In one mole of K2SO4, we have 2 mol K+. In one mole of K2CO3, we have 2 mol K+. Therefore: M of K+ ions = (0.952 M K2SO4) (2 mols K+ ions / 1 mol K2SO4) + (0.942 M K2CO3) (2 mols K+ ions / 1 mol K2CO3) = 3.79 M K+ ions
Balance the following reaction: *S2-+*Cr2O72- +** H+ à * Cr3+ + * S + * H2O
3114237
Balance the following reaction: * Br2 + * KOH à * KBrO3 + * KBr + * H2O
36153
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in SO42- → SO2 we can balance H and O by adding H2O and/or H+ if the reaction is occurring in an acidic solution. What would be the coefficient of H+ when the half reaction above is balanced?
4
Balance the following: * MnO4- + * H2O + * ClO2- à * MnO2 + * OH- + * ClO4- (basic)
423443
A hydrocarbon is a compound containing only C and H. A pure sample of a hydrocarbon was found to contain 5.261 g of C and 0.7358 g H. Which of the following is the empirical formula of the compound?
5.261 g C: 0.4380 mol C 0.7358 g H: 0.7300 mol H Divide both by the smaller one C: (0.4380/0.4380)=1 H: (0.7300/0.4380)=1.667 Since '.667' is not a whole #, we multiply both # by 3 C: (1x3)=3 H: (1.667x3)=5 Empirical formula: C3H5
A 305.1-mL sample of aqueous solution contains 214 grams of ammonium sulfate. What is the molarity of the ammonium sulfate in the solution?
5.31 ***Since we are given the mass of the solute (ammonium sulfate), we must first convert it to moles: 214 g ammonium sulfate 1.62 mol ammonium sulfate (Review lessons on molar mass if you don't know how to do this; the molar mass of ammonium sulfate is 132.2 g/mol) Since we are given the volume of the solution in mL, we need to convert it to L: 305.1 mL 0.3051 L Therefore: M=(1.62/.3051)=5.31 mol/L
Consider a half reaction where all the atoms are already balanced, as in 14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O When this half reaction is finally balanced, what would be the coefficient of e-?
6 ***Example: 14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O Step 1. Total charges: Left 14x(+1) + 1x(-2) = +12 Right 2x(+3) = +6. Step 2. Left side charge is algebraically higher. Step 3. 12-6 = 6; Add 6 electrons to left side. This will bring down total charge on left side to +6, making it equal to the right side charge.
A solution is prepared by mixing 68.7 mL of 0.219 M sodium chloride(aq) and 58.1 mL of 0.184 M sodium carbonate(aq), then adding enough water to bring the total volume to 500.0 mL. Calculate the molarity of Na+ in the resulting solution.
7.28x10^-2 M Recall that M = n/V; so to calculate moles or millimoles, we multiply M and V: millimoles sodium chloride from first solution: (0.219 M)(68.7 mL) = 15.04 mmol millimoles sodium carbonate from second solution: (0.184 M)(58.1 mL) = 10.69 mmol To get millimoles of Na+, we need to know the formulas of sodium chloride and sodium carbonate; they are NaCl and Na2CO3. There is ONE mole of Na+ in one mole of sodium chloride, but TWO moles of Na+ in one mole of sodium carbonate. Therefore: millimoles Na+ from first solution: 15.04 mmol x 1 = 15.04 mmol millimoles Na+ from second solution: 10.69 mmol x 2 = 21.38 mmol Finally, [Na+] = Molarity of Na+ = (15.04 + 21.38) mmol Na+ divided by 500.0 mL =7.28x10-2 M
How many grams of sodium sulfate are in a 70.96-mL sample of 0.922 M sodium sulfate(aq) solution?
9.29 g sodium sulfate We first need to convert our volume to liters: 70.96 mL 0.07096 L We are given that the molarity is 0.922 M or 0.922 mol/L. All we have to do now is plug in the numbers: n = (0.922 mol/L) TIMES 0.07096 L = 6.542x10-2 mol Finally, we change moles to grams: 6.542x10-2 mol 9.29 g (Note: the molar mass of sodium sulfate is 142.0 g/mol)
Which of the following is the Arrhenius definition of an acid? of an base?
A substance that yields H+ or H3O+ in water A substance that yields OH- in water
The molar mass of aluminum oxide is
Aluminum oxide is Al2O3 (2x26.98 + 3x16.00)=102 g/mol
Which of the following is the Broasted-Lowry definition of a base? of an acid?
An ion or molecule that accepts a proton An ion or molecule that donates a proton to another ion or molecule
How many moles of hydrogen atoms are in 2.73 g of hydrogen atoms?
Atomic weight of hydrogen: 1.008 g/mol (2.73 divided by 1.008)=2.71 mol hydrogen atoms
Calculate the mass of 8.09 moles of sodium atoms
Atomic weight of sodium: 22.99 g/mol (8.09x22.99)=186 g sodium
Which of the following is an empirical formula?
C2H6O Subscripts are: 2, 6, and 1
Which of the following is not an empirical formula?
C6H12O6 For an empirical formula the greatest common factor for the subscripts is must be 1
Calculate the percentage of oxygen (by mass) in CH3COOH
CH3COOH= 60.06 g O= (2x16.00)=32.00 g (32.00/60.06)x100=53.28%
CN-(aq) + H2O(l) HCN(aq) + OH-(aq) which occurs when potassium cyanide (KCN) is dissolved in water. The Bronsted-Lowry base for the forward reaction is
CN-
Calculate the percentage of Ca in CaSO4.2H2O
CaSO4.2H20= 172.2 g Ca= (1x40.08)=40.08 g (40.08/172.2)x100=23.28%
Calculate the moles of carbonate ions in a mixture of 25.2 g calcium carbonate and 12.3 g aluminum carbonate
Calcium carbonate is CaCO3, molar mass = 100.1 g/mol Aluminum carbonate is Al2(CO3)3, molar mass = 234.0 g/mol 1. Calculate moles of each compound (divide mass by molar mass): we'll get (25.2/100.1)=0.2517 mol CaCO3 (12.3/234.0)=0.05256 mol Al2(CO3)3 2. Calculate moles of carbonate in each compound: there is ONE mole of carbonate ion for every mole of CaCO3; there are THREE moles of carbonate ions for every mole of Al2(CO3)3 We'll have 0.2517 mol CO32- from CaCO3 3x0.05256 (=0.1577) mol CO32- from Al2(CO3)3 3. (0.2517+0.1577)=0.409 mol CO32-
A sample containing only calcium nitrate is found to contain 0.516 moles of nitrate ions. How many moles of calcium nitrate are in the sample?
Calcium nitrate is Ca(NO3)2 therefore: 2 nitrate ions in one formula unit of calcium nitrate (0.516 divided by 2)=0.258 mol Ca(NO3)2
Calculate the molar mass of C14H30
Carbon: 12.01x168.14=168.14 g Hydrogen: 30x1.008=30.240 g (168.14+30.240)=198.4 g/mol
Classify the reaction: LiOH + CO2 → LiHCO3
Combination
Classify the reaction that fits the following description: is a reaction involving the formation of one compound from two or more substances
Combination *** Combination: two or more reactants → one product Decomposition: One reactant → two or more products Displacement: element1 + compound1 → element2 + compound2 Double replacement: compound1 + compound2 → compound3 + compound4 Combustion: (element or compound) + oxygen → oxides of the elements
Classify the reaction: C6H12O6 + O2 → CO2 + H2O
Combustion
A pure sample of a compound is found to contain 1.113 moles of carbon atoms of and 1.855 moles hydrogen atoms. What is the empirical formula of the compound?
Divide by number by smaller the one (1.113) C: (1.113/1.113)=1 H: (1.855/1.113)=1.667 Since '.667' is not a whole # and is essentially 2/3, we can change it to a whole # by multiplying by 3 C: (1x3)=3 H: (1.667x3)=5 Empirical formula: C3H5
Classify the reaction: sodium carbonate + calcium hydroxide → sodium hydroxide + calcium carbonate
Double replacement
Classify the reaction: calcium hydroxide + sulfurous acid → calcium sulfite + water
Double replacement ***For displacement and double replacement, the compounds are either ionic, an acid, or water; the reaction can be thought of as an exchanging of partners (ions).)
A mixture of C6H6 and C7H8 is burned in excess oxygen. All of the carbon atoms in both compounds end up as part of carbon dioxide. All the hydrogen atoms in both compounds end up as part of water. If 5.851 moles of carbon dioxide, CO2, and 3.109 moles of water, H2O, are collected, how many moles of C were in the the mixture?
Every C atom from the mixture is going to give us one CO2 molecule (5.851 moles of CO2)=5.851 moles of C
True or False. The net ionic equation for the aqueous reaction between hydrofluoric acid and sodium hydroxide is H+(aq) + OH-(aq) → H2O(l)
False ***To be able to answer this type of question, you need to know: 1. the six strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4. All other acids are weak. 2. the strong, soluble (and moderately soluble) bases -- Group I hydroxides (e.g. NaOH); Group II hydroxides (except Mg(OH)2 which is insoluble). ***When you have a reaction between a strong acid and a strong base, the net ionic equation is just: H+(aq)+OH-(aq)→H2O(l).
True or False: Avogadro's number is exactly 6.02x1023
False, we do not know the exact numerical value
In the net ionic equation for a reaction between a strong acid and a strong base, which of the following are written on the reactant side?
H+ and OH-
Which of the following is the conjugate base of H2CO3?
HCO3-
Which of the following is the conjugate base of H2SO4?
HSO4-
CH4 + 2 H2O → 4 H2 + CO2 IF 7.07x103 mol CH4 is consumed as a result of this reaction, how much CO2 is produced?
IF 1x mol CH4 is consumed THEN 1x mol CO2 is produced. Amount of CH4 consumed= 1x=7.07x103 mol x=(7.07x103 mol)/1=7.07x10^3 mol Therefore: Amount of CO2 produced x=1(7.07x103)=7.07x10^3 mol
CH4 + 2 H2O → 4 H2 + CO2 How much CH4 is used up in a process where 3.64x103 mol CO2 is produced?
IF 1x mol CH4 is consumed THEN 1x mol CO2 is produced. Amount of CO2 produced= 1x=3.64x103 mol x=(3.64x103 mol)/1=3.64x10^3 mol Therefore: Amount of CH4 used up x=1(3.64x103)=3.64x10^3 mol
A pure sample of a compound is found to contain 16.62 g C and 1.744 g H. What is the empirical formula of the compound?
In this case, you are given grams. You need to remember that a gram-to-gram comparison does NOT give a relative count. You need to change grams to moles. 16.62 g C: 1.384 mol C 1.744 g H: 1.730 mol H Divide both by the smaller one C: (1.384/1.384)=1 H: (1.730/1.384)=1.250 Since '.250' is not a whole # and is essentially 1/4, we can change it to a whole # by multiplying by 4 C: (1x4)=4 H: (1.250x4)=5 Empirical formula: C4H5
What happens to the amount of solvent when an aqueous solution is diluted?
Increases **Dilution means addition of more solvent. This increases the amount of solution. Since concentration is the ratio of amount of solute to the amount of solvent or solution, then the concentration decreases upon dilution.
State whether the following is soluble or insoluble in water: PbSO4
Insoluble
State whether the following is soluble or insoluble in water: mercury(I) chloride
Insoluble ***Apply them in the order listed -- once you find a rule that applies, IGNORE the rest. Rule 1. All compounds containing sodium, potassium, ammonium, nitrate, nitrite, perchlorate, or acetate are soluble. Rule 2. Most chlorides, bromides, and iodides are soluble. Exception: insoluble if the cation is silver, mercury(I), or lead(II). Rule 3. Most sulfates are soluble. Exception: insoluble if the cation is strontium, barium, lead(II), mercury(I). Rule 4. Most hydroxides, phosphates, sulfides, chromates, and carbonates are insoluble
Which of the following salts yields a neutral solution in water?
KBr Neutral because neither the cation or the anion undergoes any significant hydrolysis
State what would happen if aqueous solutions I and II are mixed: Solution I: mercury(II) nitrate Solution II: sodium chloride
Mercury(II) chloride is soluble Mercury(I) chloride is insoluble No precipitate will be formed
Which of the following chemical reactions is an oxidation-reduction reaction?
Mg+CO2 → MgO+CO ***Mg changes from elemental zero to 2+ (oxidation process) in MgO and C changes from 4+ in CO2 to 2+ (reduction process) in CO. In all of the rest of the reactions, the oxidation state of each element remains the same for the reactants and products
A sample weighing 50.09 g contains carbon, hydrogen, and one other element. The sample is burned in excess oxygen and all the C atoms end up as part of carbon dioxide, while all the hydrogen end up as part of water. IF 40.58 g of water and 94.36 g of carbon dioxide are obtained, calculate the mass of the other element in the original sample
Molar mass of carbon dioxide: 44.01 g/mol (94.36(12.01)/44.01)=25.75 g C Molar mass of water: 18.02 g/mol (40.58(2.016)/18.02)=4.540 g H 50.09-(25.75+4.540)=19.80 g
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) which occurs when ammonium chloride (NH4Cl) is dissolved in water. The Bronsted-Lowry base for the reverse reaction is
NH3
State what would happen if aqueous solutions I and II are mixed: Solution I: AgNO3 Solution II: NaNO3
No precipitate will be formed
2 H2 + O2 2 H2O IF 12.8 g H2 and 103 g O2 were to react completely, how much H2O will be formed?
O2: 103 g:3.214 mol (you need the molar mass of O2 to do this conversion; 32.00 g/mol) H2: 12.8 g:6.355 mol (you need the molar mass of H2 to do this conversion; 2.016 g/mol 1. Figure out what x is if one reactant is consumed (in this case, let's do H2): x=(6.355 mol/2)=3.18 mol 2. Figure out what x is if the other reactant is consumed (in this case, we do O2): x=3.214 mol 3. Use the smaller x value to calculate the theoretical yield. 2x=2(3.18) mol=6.355 mol H2O Finally convert 6.355 mol H2O to grams: (6.355x18.02)=1.14x10^2 g H2O
2 H2 + O2 2 H2O IF 19.9 g of H2 and 115 g of O2 were to react completely, how much H2O will be formed?
O2: 115 g:3.606 mol (you need the molar of O2 to do this conversion, 32.00g/mol) H2: 19.9 g:9.864 mol (you need the molar mass of H2 to do this conversion, 2.016 g/mol) 1. Figure out what x is if one reactant is consumed (in this case, let's do H2): x=(9.864 mol/2)=4.932 mol 2. Figure out what x is if the other reactant is consumed (in this case, we do O2): x=3.606 mol 3. Use the smaller x value to calculate the theoretical yield. 2x=2(3.606) mol=7.212 mol H2O Finally convert 7.212 mol H2O to grams: (7.212x18.02)=129.96 g H2O
2 H2 + O2 2 H2O If a mixture of 23.0 g H2 and 187 g O2 yields 176.8 g H2O, calculate the percent yield of H2O?
O2: 187 g:5.84 mol (you need the molar mass of O2 to do this conversion; 32.00 g/mol) H2: 23.0 g:11.4 mol (you need the molar mass of H2 to do this conversion; 2.016 g/mol) 1. Figure out what x is if one reactant is consumed (in this case, let's do H2): x=(11.4 mol/2)=5.70 mol 2. Figure out what x is if the other reactant is consumed (in this case, we do O2): x=5.84 mol 3. Use the smaller x value to calculate the theoretical yield. 2x=2(5.70) mol=11.4 mol H2O Convert 11.4 mol H2O to grams: Divide actual yield (176.8 g) by theoretical yield (2.054x10^2 g) to get percent yield 86.1%
A pure sample of a compound is found to contain 1.11x1022 nitrogen atoms of and 1.10x1022 oxygen atoms. What is the empirical formula of the compound?
Obviously a 1:1 ratio Empirical formula: NO
Sodium atom becomes sodium ion
Oxidation; Sodium atom loses one electron to become Na+ ***Oxidation is a process where an atom, molecule, or ion loses electrons. Reduction is a process where electron(s) is(are) gained
State whether N is oxidized or reduced in ammonia → hydrazine
Oxidized; NH3 → N2H4 ***OXIDATION: an increase in oxidation number; implies a (fictitious or real) loss of electrons REDUCTION: a decrease in oxidation number; implies a (fictitious or real) gain of electrons
Classify the following: AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)
Precipitation reaction AgCl(s) is the precipitate ***Precipitation = formation of a solid when two solutions are mixed; one of the products must be a solid, indicated by the label (s), which is INSOLUBLE in water. Acid-Base neutralization = reaction of an acid ("hydrogen something") and a base ("something hydroxide") to form water, H2O(l) and a "salt". A "salt" is an ionic compound whose anion is not OH- nor "oxide".
Reaction 1. iron(II) nitrate + sodium carbonate Reaction 2. calcium nitrate + sodium carbonate The formula Ca2+ will
Reaction 1: Fe2+(aq)+CO32-(aq)....FeCO3(s) Reaction 2: Ca2+(aq)+CO32-(aq)....CaCO3(s) Appear in the net ionic equation for reaction 2 only
Reaction 1. silver nitrate + barium chloride Reaction 2. sodium carbonate + silver sulfate The formula SO42- will
SO42- is a spectator ion in reaction 2 Not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate
Which of the following figures best represents what happens when you mix a small amount of silver nitrate with water?
Silver nitrate is AgNO3; the cation-to-anion ratio is 1:1. It is soluble in water. To be able to answer this type of question, you need to: 1. know the names and formulas of ions 2. know how to put together a formula unit for an ionic compound (how many cations and how many anions?) 3. know your solubility rules
State whether the following is soluble or insoluble in water: Zn(C2H3O2)2
Soluble
State whether the following is soluble or insoluble in water: potassium chromate
Soluble
State whether HCl is a strong or weak acid
Strong ***The strong acids are HCl, HBr, HI, HClO4, H2SO4, and HNO3
The limiting reactant is
The limiting reactant is the reactant that would be completely consumed first.
CH4 + 2 O2 → CO2 + 2 H2O True or False: For every molecule of methane destroyed, two molecules of water are formed.
True
State whether Ba(OH)2 is a strong or weak base
Weak *** The strong bases are LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2
State whether a precipitate will be formed if the following pair of aqueous solutions are mixed: lead(II) nitrate + sodium bromide
Yes, a precipitate will be formed
State whether a precipitate will be formed if aqueous solutions containing the following ions are mixed: carbonate, sodium, magnesium, nitrate
Yes, a precipitate will be formed ***A precipitate will be formed if you have two ions that can form a compound that is insoluble in water.
Which of the following salts yields an acidic solution in water?
Zn(NO3)2 ***You must examine the ions associated with a soluble salt for possible hydrolysis to determine if the salt solution will be acidic, basic, or neutral. Cations undergo hydrolysis and produce hydronium ions (acidic) except Group I and II metals. Anions undergo hydrolysis and produce hydroxide ions (basic) except the conjugate base of a strong acid
Molarity is the ratio of which two quantities?
amount of solute in moles and liters of solution
Half reaction 1: Sn2+ Sn4+ + 2e- Half reaction 2: Fe3+ + e- Fe2+ The oxidation half reaction is ___. The reduction half reaction is ___. To construct the overall redox reaction, we multiply the oxidation half reaction by ___, multiply the reduction half reaction by ___, then combine the two half reactions. In the overall equation, after any necessary simplification, the sum of all the coefficients is ___.
half reaction 1, half reaction 2, 1, 2, 6
An aqueous solution is prepared by dissolving table salt (NaCl) in water. Which of the following calculations gives us the molarity of the solution?
moles of NaCl divided by liters of solution
