CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions)

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 1 preview image

Pay it forward.MODULE 1 EXAMQuestion 1Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.1.Convert 845.3 to exponential form and explain your answer.2.Convert 3.21 x 10-5to ordinary form and explain your answer.1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places= 8.453 x 1022.Convert 3.21 x 10-5= negative exponent = smaller than 1, move decimal 5places = 0.0000321Question 2Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Using the following information, do the conversions shown below, showing allwork:1 ft = 12 inches1 pound = 16 oz1 gallon = 4 quarts1 mile = 5280 feet1 ton = 2000 pounds1 quart = 2 pintskilo (= 1000)milli (= 1/1000)centi (=1/100)deci (= 1/10)1.24.6 grams = ? kg2.6.3 ft = ? inches1.24.6 grams x 1 kg / 1000 g = 0.0246 kg2.6.3 ft x 12 in / 1 ft = 75.6 inchesplease always use the correct units in your final answerQuestion 3

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 2 preview image

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 3 preview image

Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Do the conversions shown below, showing all work:1.28oC = ?oK2.158oF = ?oC3.343oK = ?oF1.28oC + 273 = 301oKoC→oK (make larger)+2732.158oF - 32 ÷ 1.8 = 70oCoF→oC (make smaller)-32 ÷1.83.343oK - 273 = 70oC x 1.8 + 32 = 158oFoK→oC→oFQuestion 4Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Be sure to show the correct number of significant figures in each calculation.1.Show the calculation of the mass of a 18.6 ml sample of freon withdensity of 1.49 g/ml2.Show the calculation of the density of crude oil if 26.3 g occupies 30.5ml.1.M = D x V = 1.49 x 18.6 = 27.7 g2.D = M / V = 26.3 / 30.5 = 0.862 g/mlQuestion 5Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.1.3.0600 contains ? significant figures.2.0.0151 contains ? significant figures.

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 4 preview image

3.3.0600 ÷ 0.0151 = ? (give answer to correct number of significantfigures)1.3.0600contains5significant figures.2.0.0151contains3significant figures.3.3.0600 ÷ 0.0151 = 202.649 = 203 (to 3 significant figures for 0.0151)Question 6Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Classify each of the following as an element, compound, solution orheterogeneous mixture and explain your answer.1.Coca cola2.Calcium3.Chili1.Coca cola - is not on periodic table (not element) - no element names(not compound)appears to be one substance = Solution2.Calcium - is on periodic table = Element3.Chili - is not on periodic table (not element) - no element names (notcompound)appears as more than one substance (meat, beans,sauce) = Hetero MixQuestion 7Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Classify each of the following as a chemical change or a physical change1.Charcoal burns2.Mixing cake batter with water

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 5 preview image

3.Baking the batter to a cake1.Charcoal burns - burning always = chemical change2.Mixing cake batter with water - mixing = physical change3.Baking the batter to a cake - baking converts batter to new material =chemical changeQuestion 8Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the full Nuclear symbol including any + or - charge (n), the atomicnumber (y), the mass number (x) and the correct element symbol (Z) foreach element for which the protons, neutrons and electrons are shown -symbol should appear as follows:xZy+/- n31 protons, 39 neutrons, 28 electrons31 protons = Ga31, 39 neutrons =70Ga31, 28 electrons = (+31 - 28 = +3)=70Ga31+3Question 9Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Name each of the following chemical compounds.Be sure to name all acidsas acids (NOT for instance as binary compounds)1.PF52.Al2(CO3)33.H2CrO41.PF5- binary molecular = phosphorus pentafluoride2.Al2(CO3)3- nonbinary ionic = aluminum carbonate

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 6 preview image

3.H2CrO4- nonbinary acid = chromic acidincorrect fluoride prefixQuestion 10Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Write the formula for each of the following chemical compounds explainingthe answer with appropriate charges and/or prefixes and/or suffixes.1.Carbon monoxide2.Manganese (IV) acetate3.Phosphorous acid1.Carbon monoxide - ide = binary, mono = 1 O= CO2.Manganese (IV) acetate - Mn+4, C2H3O2-1= Mn(C2H3O2)43.Phosphorous acid - nonbinary acid of H + phosphite (PO3-3) = H3PO3MODULE 2 EXAMQuestion 1Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the calculationof the molecular weight for the followingcompounds, reporting your answer to 2 places after the decimal.1.Al2(CO3)32.C8H6NO4Cl1.2Al + 3C + 9O = 233.99

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 7 preview image

2.8C + 6H + N + 4O + Cl = 215.59Question 2Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the calculationof the number of moles in the given amount of thefollowing substances.Report your answerto 3 significant figures.1.13.0 grams of (NH4)2CO32.16.0 grams of C8H6NO4Br1.Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole2.Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 moleQuestion 3Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the calculationof the number of grams in the given amount of thefollowing substances.Report your answer to 1 place after the decimal.1.1.20 moles of (NH4)2CO32.1.04 moles of C8H6NO4Br1.Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams2.Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 gramsQuestion 4Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 8 preview image

Show the calculation of the percent of each element present in the followingcompounds.Report your answer to 2 places after the decimal.1.(NH4)2CrO42.C8H8NOI1.%N = 2 x 14.01/152.08 x 100 = 18.43%%H = 8 x 1.008/152.08x 100 = 5.30%%Cr = 1 x 52.00/152.08 x 100 = 34.20%%O = 4 x16.00/152.08 x 100 = 42.08%2.%C = 8 x 12.01/261.05 x 100 = 36.80%%H = 8 x 1.008/261.05x 100 = 3.09%%N = 1 x 14.01/261.05 x 100 = 5.37%%O = 1 x16.00/261.05 x 100 = 6.13%%I = 1 x 126.9/261.05 x 100 = 48.61%Question 5Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the calculation of the empirical formula for each compound whoseelemental composition is shown below.38.76% Ca, 19.87% P, 41.27% O38.76% Ca / 40.08 = 0.9671 / 0.6416 = 1.5 x 2 = 319.87% P / 30.97 =0.6416/ 0.6416 = 1 x 2 = 241.27% O / 16.00 = 2.579 / 0.6416 = 4 x 2 = 8→Ca3P2O8Question 6

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 9 preview image

Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Balance each of the following equationsby placing coefficients in front ofeach substance.1.C6H6+O2→CO2+H2O2.As+O2→As2O53.Al2(SO4)3+Ca(OH)2→Al(OH)3+CaSO41.2 C6H6+15 O2→12 CO2+6 H2O2.4 As+5 O2→2 As2O53.Al2(SO4)3+3 Ca(OH)2→2 Al(OH)3+3 CaSO4Question 7Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Classify each of the following reactions as either:CombinationDecompositionCombustionDouble ReplacementSingle Replacement1.H2SO4→SO3+H2O2.S+3 F2→SF63.H2+NiO→Ni+H2O1.H2SO4→SO3+H2O = Decomposition, One reactant→Two Products2.S+3 F2→SF6= Combination. Two reactants→One product

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 10 preview image

3.H2+NiO→Ni+H2O = Single Replacement, Hydrogen displacesmetal ionQuestion 8Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the calculation of the oxidation number (charge)of ONLY the atomswhich are changing in the following redox equations.Na2HAsO3+KBrO3+HCl→NaCl+KBr+H3AsO4Na2HAsO3+KBrO3+HCl→NaCl+KBr+H3AsO4Na2HAsO3: Na is metal in group I = +1 (total is +2), H = +1, each O is -2(total is -6), soAs is +3H3AsO4: H is +1 (total is +3), each O is -2 (total is -8), soAs is +5KBrO3: K is metal in group I = +1, each O is -2 (total is -6), soBr is +5KBr: K is metal in group I = +1, soBr is -1Question 9Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the balancing of the following redox equation, including thedetermination of the oxidation number (charge) of ONLY the atoms which arechanging.KMnO4+KI+H2O→KIO3+MnO2+KOHMn compounds x 2 ; I compounds x 1 =2KMnO4 + 1 KI + 1 H2O→1 KIO3+ 2MnO2+ 2KOHKMnO4+KI+H2O→KIO3+MnO2+KOHKMnO4: K is metal in group I = +1, each O is -2 (total is -8), soMn is +7

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CHEM103 Portage Learning Module 1 to 6 Exam with Answers (72 Solved Questions) - Page 11 preview image

MnO2: Each O is -2 (total is -4), soMn is +4KI: K is metal in group I = +1, soI is -1KIO3: K is metal in group I = +1, each O is -2 (total is -6), soI is +5Since Mn (on left side) is +7 and Mn (on right side) is +4: Mn changes by 3Since I (on left side) is -1 and I (on right side) is +5: I changes by 6Multiply Mn compounds by 2 and I compounds by 1 and after balancing otheratoms =2 KMnO4+1 KI+1 H2O→1 KIO3+2 MnO2+2 KOHQuestion 10Click this link to access thePeriodic Table.This may be helpful throughoutthe exam.Show the balanced equation and the calculation of the number of moles andgrams of CO2formed from 20.6 grams of C6H6.Show your answers to 3significant figures.C6H6+O2→CO2+H2OMolar mass of CO2 = 44.01 g/molmass of CO2 = 1.5822 ml x 44.01 g/mol =69.63 gramsmass of CO2 = 69.63 gramsmoles of CO2 = 1.58 mole2 C6H6+15 O2→12 CO2+6 H2O20.6 g / (6 x 12.01 + 6 x 1.008) = 20.6 / 78.108 = 0.2637 mole x 12/2=1.58 mole CO21.582 mole CO2x (12.01 + 2 x 16.00) =69.6 g CO2

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