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Chem 103 Module 1 to 6 Exam Question and answers
(Portage learning) 2025/
MODULE 1 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- Convert 845.3 to exponential form and explain your answer.
- Convert 3.21 x 10 -^5 to ordinary form and explain your answer.
- Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 102
- Convert 3.21 x 10 -^5 = negative exponent = smaller than 1, move decimal 5 places = 0.
Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam. Using the following information, do the conversions shown below, showing all work: 1 ft = 12 inches 1 mile = 5280 feet 1 pound = 16 oz 1 ton = 2000 pounds 1 gallon = 4 quarts 1 quart = 2 pints kilo (= 1000) 1/100) milli (= 1/1000) deci (= 1/10) centi (=
- 24.6 grams =? kg
- 6.3 ft =? inches
- 24.6 grams x 1 kg / 1000 g = 0.0246 kg
- 6.3 ft x 12 in / 1 ft = 75.6 inches please always use the correct units in your final answer
Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work:
- 28 oC =? oK
- 158 oF =? oC
- 343 oK =? oF
- 28 oC + 273 = 301 oK oC → oK (make larger)
- 158 oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller)
- 343 oK - 273 = 70 oC x 1.8 + 32 = 158 oF oK → oC → oF
Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam. Be sure to show the correct number of significant figures in each calculation.
- Show the calculation of the mass of a 18.6 ml sample of freon with density of 1.49 g/ml
- Show the calculation of the density of crude oil if 26.3 g occupies 30. ml.
- M = D x V = 1.49 x 18.6 = 27.7 g
- D = M / V = 26.3 / 30.5 = 0.862 g/ml
Question 5
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- 3.0600 contains? significant figures.
- 0.0151 contains? significant figures.
31
Question 8
- Baking the batter to a cake
- Charcoal burns - burning always = chemical change
- Mixing cake batter with water - mixing = physical change
- Baking the batter to a cake - baking converts batter to new material = chemical change Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the full Nuclear symbol including any + or - charge (n), the atomic number (y), the mass number (x) and the correct element symbol (Z) for each element for which the protons, neutrons and electrons are shown - symbol should appear as follows: x Z y+/-^ n 31 protons, 39 neutrons, 28 electrons 31 protons = Ga 31 , 39 neutrons = 70 Ga 31 , 28 electrons = (+31 - 28 = +3) = 70 Ga +
Question 9
Click this link to access the Periodic Table. This may be helpful throughout the exam. Name each of the following chemical compounds. Be sure to name all acids as acids (NOT for instance as binary compounds)
- PF 5
- Al 2 (CO 3 ) 3
- H 2 CrO 4
- PF 5 - binary molecular = phosphorus pentafluoride
- Al 2 (CO 3 ) 3 - nonbinary ionic = aluminum carbonate
2 3 3 3
- H 2 CrO 4 - nonbinary acid = chromic acid incorrect fluoride prefix
Question 10
Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the formula for each of the following chemical compounds explaining the answer with appropriate charges and/or prefixes and/or suffixes.
- Carbon monoxide
- Manganese (IV) acetate
- Phosphorous acid
- Carbon monoxide - ide = binary, mono = 1 O = CO
- Manganese (IV) acetate - Mn+4, C 2 H 3 O -^1 = Mn(C 2 H 3 O 2 ) 4
- Phosphorous acid - nonbinary acid of H + phosphite (PO -^3 ) = H PO MODULE 2 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal.
- Al 2 (CO 3 ) 3
- C 8 H 6 NO 4 Cl
- 2Al + 3C + 9O = 233.
Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal.
- (NH 4 ) 2 CrO 4
- C 8 H 8 NOI
- %N = 2 x 14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152. x 100 = 5.30% %Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x 16.00/152.08 x 100 = 42.08%
- %C = 8 x 12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261. x 100 = 3.09% %N = 1 x 14.01/261.05 x 100 = 5.37% %O = 1 x 16.00/261.05 x 100 = 6.13% %I = 1 x 126.9/261.05 x 100 = 48.61%
Question 5
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 38.76% Ca, 19.87% P, 41.27% O 38.76% Ca / 40.08 = 0.9671 / 0.6416 = 1.5 x 2 = 3 19.87% P / 30.97 = 0.6416 / 0.6416 = 1 x 2 = 2 41.27% O / 16.00 = 2.579 / 0.6416 = 4 x 2 = 8 → Ca 3 P 2 O 8
Question 6
Click this link to access the Periodic Table. This may be helpful throughout the exam. Balance each of the following equations by placing coefficients in front of each substance.
- C 6 H 6 + O 2 → CO 2 + H 2 O
- As + O 2 → As 2 O 5
- Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4
- 2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O
- 4 As + 5 O 2 → 2 As 2 O 5
- Al 2 (SO 4 ) 3 + 3 Ca(OH) 2 → 2 Al(OH) 3 + 3 CaSO 4
Question 7
Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following reactions as either: Combination Decomposition Combustion Double Replacement Single Replacement
- H 2 SO 4 → SO 3 + H 2 O
- S + 3 F 2 → SF 6
- H 2 + NiO → Ni + H 2 O
- H 2 SO 4 → SO 3 + H 2 O = Decomposition, One reactant → Two Products
- S + 3 F 2 → SF 6 = Combination. Two reactants→ One product
MnO 2 : Each O is - 2 (total is - 4), so Mn is + KI: K is metal in group I = +1, so I is - 1 KIO 3 : K is metal in group I = +1, each O is - 2 (total is - 6), so I is + Since Mn (on left side) is +7 and Mn (on right side) is +4: Mn changes by 3 Since I (on left side) is - 1 and I (on right side) is +5: I changes by 6 Multiply Mn compounds by 2 and I compounds by 1 and after balancing other atoms = 2 KMnO 4 + 1 KI + 1 H 2 O → 1 KIO 3 + 2 MnO 2 + 2 KOH
Question 10
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balanced equation and the calculation of the number of moles and grams of CO 2 formed from 20.6 grams of C 6 H 6. Show your answers to 3 significant figures. C 6 H 6 + O 2 → CO 2 + H 2 O Molar mass of CO2 = 44.01 g/mol mass of CO2 = 1.5822 ml x 44.01 g/mol =69.63 grams mass of CO2 = 69.63 grams moles of CO2 = 1.58 mole 2 C 6 H 6 + 15 O 2 → 12 CO 2 + 6 H 2 O 20.6 g / (6 x 12.01 + 6 x 1.008) = 20.6 / 78.108 = 0.2637 mole x 12/ = 1.58 mole CO 2 1.582 mole CO 2 x (12.01 + 2 x 16.00) = 69.6 g CO 2
MODULE 3 EXAM
Click this link to access the Periodic Table. This may be helpful throughout the exam. A reaction between HCl and NaOH is being studied in a styrofoam coffee cup with NO lid and the heat given off is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed.
- The type of thermochemical process
- The amount of heat released in the reaction of HCl with NaOH
- Heat given off = Exothermic process
- The amount of heat released = Heat of reaction
Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- Show the calculation of the final temperature of the mixture when a 22.8 gram sample of water at 74.6oC is added to a 14.3 gram sample of water at 24.3oC in a coffee cup calorimeter. c (water) = 4.184 J/g oC
- Show the calculation of the energy involved in freezing 54.3 grams of water at 0oC if the Heat of Fusion for water is 0.334 kJ/g
- (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool H2O)
Question 1
f 4 f f 2 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 3 C (graphite) + 4 H 2 (g) → C 3 H 8 (g) by using the following thermochemical data: C (graphite) kJ
- O 2 (g)^ → CO 2 (g)^ ΔH = - 393. 2 H 2 (s) kJ
- O 2 (g) → 2 H 2 O(l) ΔH = - 571. C 3 H 8 (g) (^) + 5 O 2 (g) (^) → 3 CO 2 (g) (^) + 4 H 2 O(l) (^) ΔH = - 2220.0 kJ Your Answer: 3 (C (graphite) + O 2 (g) → CO 2 (g) ΔH = - 393.51 kJ) 2 (2 H 2 (s) + O 2 (g) → 2 H 2 O(l) ΔH =
- 571.66 kJ) 3 CO 2 (g) + 4 H 2 O(l) → C 3 H 8 (g) + 5 O 2 (g) ΔH = + 2220.0 kJ 3 C (graphite) + 4 H 2 (g) → C 3 H 8 (g) **ΔHrxn =
- 103.85 kJ** ΔHrxn = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.
Question 5
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 CH 4 (g) + 3 O 2 (g) → 2 CO (g) + 4 H 2 O (l) by using the following ΔHf^0 data: ΔH 0 CH (g) = - 74.6 kJ/mole, ΔH 0 CO (g) = - 110.5 kJ/mole, ΔH 0 H O (l) = - 285. kJ/mole 2 CH 4 (g) + 3 O 2 (g) → 2 CO (g) + 4 H 2 O (l) ΔHf^0 CH 4 (g) = - 74.6 kJ/mole, ΔHf^0 CO (g) = - 110.5 kJ/mole, ΔHf^0 H 2 O (l) = - 285. kJ/mole
ΔHrxn = 2(+74.6) + 3(0) + 2(-110.5) + 4(-285.8) = - 1215.0 kJ/mole
Question 6
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the new temperature of a gas sample has an original volume of 740 ml when collected at 710 mm and 35oC when the volume becomes 460 ml at 1.20 atm. P1V1/T1 = P2V2/T T2 = 1.2 x 460x308 / 0.934 x 740 = 245.98 kelvin = - 27.17oC (Pi x Vi ) / Ti = (Pf x Vf ) / Tf 740 ml/1000 = 0.740 liters = Vi 710 mm/760 = 0.934 atm = Pi 460 ml/1000 = 0.460 liters = Vf 1.20 atm = Pf 35 oC + 273 = 308 oK = Ti (0.934) x (0.740) / 308 = (1.20) x (0.460) / Tf Tf = 246 oK
Question 7
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume occupied by a gas sample containing 0.632 mole collected at 710 mm and 35oC. P x V = n x R x T
Question 10
Question 2
Question 3
XN2 = 0.2712 / (0.2712 + 0.2625) = 0.
XO2 = 0.2625 / (0.2712 + 0.2625) = 0.
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight of an unknown gas if the rate of effusion of carbon dioxide gas (CO 2 ) is 1.83 times faster than that of an unknown gas. (rN2 /runknown)^2 = MWunknown / MWCO (1.83/1)^2 = MWunknown / 44.01 MWunknown = (1.83)^2 x 44. = 147. calculation error; minus 2. MODULE 4 EXAM Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s^2 2s^2 , etc.) for the Fe 26 atom. Fe 26 = 26 electrons = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s^2 2s^2 , etc.) for the S 16 atom. S 16 = 16 electrons = 1s^2 2s^2 2p^6 3s^2 3p^4 Click this link to access the Periodic Table. This may be helpful throughout the exam.
Question 1
Write the subshell electron configuration (i.e.1s^2 2s^2 , etc.) for the P 15 atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are. P 15 = 15 electrons = 1s^2 2s^2 2p^6 3s^2 3p^3 = 5 valence electrons
Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam. *** For the following question, use the "Insert Math Equation" tool (indicated by the x icon on the toolbar and then choose arrows from the window which opens).** Using up and down arrows, write the orbital diagram for the Ti 22 atom. Ti 22 = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
Question 5
Click this link to access the Periodic Table. This may be helpful throughout the exam. *** For the following question, use the "Insert Math Equation" tool (indicated by the x icon on the toolbar and then choose arrows from the window which opens).** Using up and down arrows, write the orbital diagram for the V 23 atom and identify which are unpaired electrons and determine how many unpaired electrons there are. V 23 = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ = 3 unpaired electrons
Question 8
Click this link to access the Periodic Table. This may be helpful throughout the exam.
- List and explain which of the following atoms forms a positive ion with more difficulty. Sn or I
- List and explain which of the following is the smaller atom. Sn or Te
- I forms a positive ion less easily than Sn since ionization potential increases as you go to the right in a period which means that I with the higher ionization potential requires more energy to lose an electron and form a positive ion so it does so less easily.
- Te is smaller than Sn since atomic size decreases as you go to the right in a period which means that Te which is further to the right is smaller.
Question 9
Click this link to access the Periodic Table. This may be helpful throughout the exam. On a piece of scratch paper, draw the orbital configuration of the C 6 atom and use it to draw the Lewis structure for the C 6 atom. Then choose the correct Lewis structure for C 6 from the options listed below. C.
Question 10
Click this link to access the Periodic Table. This may be helpful throughout the exam. On a piece of scratch paper, draw the orbital configuration of the As 33 atom and use it to draw the Lewis structure for the As 33 atom. Then choose the correct Lewis structure for the As 33 from the options listed below. Your Answer: As33 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^3 ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓↑↓ ↑↑↑ B MODULE 5 EXAM
Question 1
Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the determination of the charge on the ion formed by the Se 34 atom. Se 34 (nonmetal = gain electrons) 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4 gain 2e → Se-^2
Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam. H = 2. Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.
