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Portage Learning / CHEM 103 / Exam 1-6 Final
Updated 2022-2023, Exams for Chemistry
EXAM 1
Question 1
- Convert 1005.3 to exponential form and explain your answer.
- Convert 4.87 x 10-6 to ordinary form and explain your answer. Answer:
- Convert 1005.3 = larger than 1 = positive exponent, move decimal 3 places = 1.0053 x 103
- Convert 4.87 x 10-6 =negative exponent =smaller than 1, move decimal 6 places = 0. Question 2 Using the following information, do the conversions shown below, showing all work:
2023 Portage Learning CHEM 103 Exam 1-
Final
LATEST
Exam 1- 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts
Portage Learning / CHEM 103 / Exam 1-6 Final
Updated 2022-2023, Exams for Chemistry
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints
- 2.73 liters x 1000 ml / 1 liter = 2730 ml
- 8.6 pts x 1 qt / 2 pts = 4.3 qts Question 3 Do the conversions shown below, showing all work:
- 248oC =? oK
- 25oF =? oC
- 175oK =? oF deci (= 1/10)
- 2.73 liters =? ml 2. 8.6 pts =? qts Answer: Answer:
- 248oC + 273 = 521 oK oC → oK (make larger) +
- Show the calculation of the mass of a 17.9 ml sample of chloroform with density of 1.49 g/ml 2.Show the calculation of the volume of 19.4 grams of cresol with density of 1.02 g/ml Answer:
- M = D x V = 1.49 x 17.9 = 26.7 g
- V = M / D = 19.4 / 1.02 = 19.0 ml Question 5
- 1.35601 contains? significant figures.
- 0.151 contains? significant figures.
- 25oF - 32 ÷ 1.8 = -3.88 oC oF → oC (make smaller) - ÷1.
- 175oK^ -^273 =^ -98^ oC^ x^ 1.8^ +^32 =^ - 144.4 oF oK → oC → oF Question 4 Be sure to show the correct number of significant figures in each calculation.
- 1.35601 + 0.151 =? (give answer to correct number of significant figures) Answer:
compound) appears to be one substance = Solution
Question 7 Classify each of the following as a chemical change or a physical change
- A silver spoon forms a black tarnish coating
- Food is digested
- Rain freezes on a road on a very cold day Answer:
- Silver spoon forms black coating - this is Ag → Ag2S (color change) = chemical change
- Food is digested - breakdown of carbs, proteins, fats to new materials = chemical change
- Rain freezes on a road on a very cold day - freezing = physical change Question 8 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: xZy+/- n 53 protons, 74 neutrons, 54 electrons
- Diiodine pentoxide
- Potassium phosphide
- Iron (III) cyanide - Fe+3, CN-1 = Fe(CN)
- Diiodine pentoxide - ide = binary, two I, 5 O = I2O
- Potassium phosphide - ide = binary K+1, P-3 = K3P
EXAM 2
Question 1 Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal.
- (NH4)2CrO
Question 2 Show the calculation of the number of moles in the given amount of the following substances.
- 12.0 grams of (NH4)2CrO
- 15.0 grams of C8H8NOI Answer:
- Moles = grams / molecular weight = 12.0 / 152.08 = 0.0789 mole
- Moles = grams / molecular weight = 15.0 / 261.05 = 0.0575 mole
2. C8H8NOI
Answer:
- 2N + 8H + Cr + 4O = 152.
- 8C + 8H + N + O + I = 261. Question 3 Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal.
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 Answer: 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 → Ca3P2O8 Question 6 Balance each of the following equations by placing coefficients in front of each substance.
- C4H10 + O2 → CO2 + H2O
- P + O2 → P2O
- Al + H2SO4 → Al2(SO4)3 + H2 Answer:
- 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
- 4 P + 5 O2 → 2 P2O
- 2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2 Question 7 Classify each of the following reactions as either: Combination Decomposition
Combustion Double Replacement Single Replacement
- C5H12 + 8 O2 → 5 CO2 + 6 H2O
- Zn + CuSO4 → Cu + ZnSO
- 2 Fe + 3 Cl2 → 2 FeCl3 Answer:
- C5H12 + 8 O2 → 5 CO2 + 6 H2O = Combustion, Hydrocarbon
- Zn + CuSO4 → Cu + ZnSO4 = Single Replacement, Metal displaces metal ion
- 2 Fe + 3 Cl2 → 2 FeCl3 = Combination. Two reactants→ One product Question 8 Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations. Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO Answer: Na2HAsO3: Na is metal in group I = +1 (total is +2), H = +1, each O is - (total is -6), so As is + H3AsO4: H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO3: K is metal in group I = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group I = +1, so Br is -
Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 25.4 grams of C8H18. Show your answers to 3 significant figures. C8H18 + O2 → CO2 + H2O Answer: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O 25.4 g / (8 x 12.01 + 18 x 1.008) = 25.4 / 114.224 = 0.2223 mole x 16/2 = 1. mole CO 1.78 mole CO2 x (12.01 + 2 x 16.00) = 78.3 g CO
Exam 3
Question 1
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.
1. The type of thermochemical process
2. The amount of heat released in the reaction of HCl with NaOH
Answer:
1. Heat given off = Exothermic process
2. The amount of heat released = Heat of reaction
2. ql↔g = m x ∆Hvapor = 95.6 g x 2.26 kJ/g = 216.1 kJ (since
heat is removed) = - 216.1 kJ
Question 3
Show the calculation of the amount of heat involved if 18.3 g of S is
reacted with excess O 2 to yield sulfur trioxide by the following reaction
equation. Report your answer to 4 significant figures.
2 S (s) + 3 O 2 (g) → 2 SO 3 (g) ΔH = - 792 kJ
Answer:
2 S (s) + 3 O 2 (g) → 2 SO 3 (g) ΔH = - 792
kJ ΔHrx is for 2 mole of S
reaction uses 18.3 g S = 18.3/32.07 = 0.
mole S q = ΔHrx x new moles / original moles
q = -792 kJ x 0.5706 mole S / 2 mole S = 226.0 kJ given off
Question 4
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) + O 2 (g) → CO 2 (g) ΔH = -
393.51 kJ
2 H 2 (s)
571.66 kJ
+ O 2 (g) → 2 H 2 O(l) ΔH = -
C 3 H 8 (g)
2220.0 kJ
Answer:
+ 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l) ΔH = -
3 (C (graphite) +^ O^2 (g)^ →^ CO^2 (g)^ ΔH = - 393.
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.
kJ3 C (graphite) + 4 H 2 (g) → C 3 H 8 (g) ΔHrxn =
ΔHrxn = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.85kJ
Question 5
Show the calculation of the heat of reaction (ΔHrxn) for the reaction:
2 C 6 H 6 (g) + 9 O 2 (g) → 12 CO (g) + 6 H 2 O (l)
by using the following thermochemical data:
