Test Bank Chapter 3 Composition of Substances and Solutions - Chemistry 2e Complete Test Bank by Paul Flowers. DOCX document preview.
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Chapter 03: Composition of Substances and Solutions
- Which of the following compounds contains the largest number of atoms? (Outcome # 1,2) DOK 2
- 1.00 mole of NH3
- 1.00 mole of H2SO4
- 1.00 mole of HBr
- What is the molar mass of (NH4)3PO4? (Outcome # 1,2) DOK 2
- 113 g
- 121 g
- 149 g
- What is the mass in grams of 6.022 x 1023 molecules of CO2? (Outcome # 1,2) DOK 3
- 22 g
- 44 g
- 66 g
- What is the total number of atoms contained in 2.00 moles of iron? (Outcome # 1,2) DOK 2
- 118
- 6.02 x 1023
- 1.2 x 1024
- What is the molar mass of K2CO3? (Outcome # 1,2) DOK 2
- 138 g/mol
- 106 g/mol
- 99 g/mol
- What is the total number of atoms contained in 1.00 mole of sodium? (Outcome # 1,2,9) DOK 2
- 118
- 6.02 x 1023
- 1.2 x 1024
- If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molarity of the HCl solution. (Outcome # 1,2,5,8,9,10) (DOK 3)
- 6.07 M
- 0.22 M
- 14.7 M
- To dilute a HCl solution from 0.500 M to 0.100 M the final volume must be: (Outcome # 1,2,4,5,8) DOK 2
- 1/5 the original volume
- 5 times the original volume
- 50 L
- How would you prepare a 1 M solution of sodium hydroxide? (Outcome # 5) (DOK 3)
- add 40.0 grams of sodium hydroxide to 1 L of water
- add 40.0 grams of sodium hydroxide to 100 L of water
- add 40.0 grams of sodium hydroxide to 10 L of water
- How would you prepare a 0.5 M solution of sodium hydroxide? (Outcome # 5) (DOK 3) (Paired item 1)
- add 40.0 grams of sodium hydroxide to 1 L of water
- add 20.0 grams of sodium hydroxide to 1 L of water
- add 20.0 grams of sodium hydroxide to 10 L of water
- An experiment requires 15.0 mL of a 0.100 M solution of HCl. If you have a 12.0 M HCl stock solution in the laboratory, how would you prepare enough 0.1 M solution to run at least 3 experiments? (Outcome # 5) (DOK 3)
- add 0.833 mL 12 M HCl stock solution to 99.167 mL water
- add 0.833 mL 12 M HCl stock solution to 49.167 mL water
- add 0.833 mL 12 M HCl stock solution to 199.167 mL water
- What is the mass in grams of 1.59 moles of aluminum oxide? (Outcome # 2) (DOK 2)
- 162 g
- 0.0156 g
- 64.1 g
- What is the mass in grams of 0.852 moles of ferric oxide? (Outcome # 2) (DOK 3) (Paired item 1)
- 5.34 x 10-3 g
- 187 g
- 136 g
- How many moles are in 1.77 kg of cuprous sulfate? (Outcome # 2) (DOK 3)
- 7.93 x 10-3 moles
- 7.93 moles
- 395 moles
- How many grams are in 2.05 x 1023 molecules of dinitrogen pentoxide? (Outcome # 2) (DOK 3)
- 317 g
- 36.8 g
- 0.00315 g
- How many grams are in 1.02 x 1022 molecules of nitric acid? (Outcome # 2) (DOK 3) (Paired item 1)
- 1.07 g
- 3720 g
- 2.69 x 10-4 g
- What is the total number of atoms contained in 2.11 moles of nickel? (Outcome # 2) (DOK 2)
- 2.85 x 1023 atoms
- 1.27 x 1024 atoms
- 3.50 x 10-24 atoms
- How many atoms are in 1.00 mole of Au? (Outcome # 2) (DOK 2) (Paired item 1)
- 6.02 x 1023 atoms
- 1.66 x 10-24 atoms
- 197 atoms
- A 44.8 g rhodium sample contains how many rhodium atoms? (Outcome # 2) (DOK 3)
- 2.62 x 1023 atoms
- 6.02 x 1023 atoms
- 2.70 x 1025 atoms
- What is the molar mass of potassium nitrate? (Outcome # 2) (DOK 2)
- 85.10 g/mol
- 140.20 g/mol
- 101.10 g/mol
- Commercial grade fuming nitric acid contains about 90% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the HNO3 solution. (Outcome # 9) (DOK 3)
- 0.214 M
- 21.4 M
- 2.14 M
- Which of the following is a method used to allow scientists to determine an empirical formula and identify an organic compound by combusting a sample and the resulting combustion products are quantitatively analyzed? (Outcome # 7) (DOK 1)
- gravimetric analysis
- combustion analysis
- thermal analysis
- A 3.00 g sample of unknown hydrocarbon is prepared for combustion analysis. After the hydrocarbon undergoes complete combustion, 8.80 g of CO2 and 5.40 g of H2O are produced. What is the empirical formula of the unknown? (Outcome # 7) (DOK 3)
- CH3
- CH
- CH2
- A 0.125 g sample of unknown hydrocarbon is prepared for combustion analysis. After the hydrocarbon undergoes complete combustion, 0.4225 g of CO2 and 0.0865 g of H2O are produced. What is the empirical formula of the unknown? (Outcome # 7) (DOK 3) (Paired item 1)
- CH3
- CH
- CH2
- An unknown organic compound (0.315 g) containing only C, H, and O produces 0.771 of CO2 and 0.105 g of H2O when it undergoes complete combustion. The approximate molar mass is 108 g/mol. Which of the following compounds could be the identification of the unknown? (Outcome # 7) (DOK 3)
- para-Cresol, CH3C6H4(OH)
- Benzyl alcohol, C6H5CH2OH
- 1,2-Benzoquinone, C6H4O2
- Which of the following is an experimental technique used to determine the mass percent of elements found in a compound? (Outcome # 7) (DOK 1)
- gravimetric analysis
- mass analysis
- elemental analysis
- An unknown hydrocarbon compound was analyzed for hydrogen by elemental analysis and results show that it contains 15.88 % H. What is the empirical formula? (Outcome # 7) (DOK 2)
- C4H9
- C8H18
- C12H27
- A 5.91 g unknown sample analyzed by elemental analysis revealed a composition of 37.51 % C. In addition, it was determined that the sample contains 1.4830 x 1023 hydrogen atoms and 1.2966 x 1023 oxygen atoms. What is the empirical formula? (Outcome # 7) (DOK 3)
- C6H8O7
- C5H6O3
- C8H16O5
- What is the concentration in ppm of a 3.7 L solution (d = 1.00 g/mL) containing 4.21 x 10-7 kg of the pesticide DDT? (Outcome # 8) (DOK 2)
- 1.14 x 10-1 ppm
- 1.14 x 102 ppm
- 1.14 x 101 ppm
- When a salt is dissolved in water, it is represented by the following notation in a chemical reaction? (Outcome # 6) (DOK 1)
- (aq)
- (l)
- (s)
- How is a solid phase represented in a chemical reaction? (Outcome # 6) (DOK 1)
- (s)
- (g)
- (l)
- How is a liquid phase represented in a chemical reaction? (Outcome # 6) (DOK 1)
- (s)
- (g)
- (l)
- How is a gas phase represented in a chemical reaction? (Outcome # 6) (DOK 1)
- (s)
- (g)
- (l)
- What are the proper units for molar mass in chemistry? (Outcome # 6) (DOK 1)
- amu
- g/mol
- g/mmol
- How many atoms can be found in a mole of mercury? (Outcome # 2) (DOK 2)
- 201 atoms
- 6.02 x 1023 atoms
- 1.21 x 1026 atoms
- A 21.8 g sample of water contains how many water molecules? (Outcome # 2) (DOK 3)
- 7.29 x 1023 molecules
- 6.02 x 1023 molecules
- 2.01 x 10-24 molecules
- A sample has a molar mass of 143 g/mol, identify the sample. (Outcome # 2) (DOK 1)
- AuCl
- AcCl
- AgCl
- What is the molar mass of 2.50 g of an unknown sample dissolved in 130. mL of water with a concentration of 0.480 M? (Outcome # 4) (DOK 3)
- 0.062 g/mol
- 8.65 g/mol
- 40.1 g/mol
- 2.50 g of an unknown base is dissolved in 289 mL of water and has a concentration of 0.216 M. What is the identification of the unknown base? (Outcome # 4) (DOK 3)
- NaOH
- LiOH
- KOH
- How many grams of Hydrogen are present in three moles of Hydrogen? (Outcome # 1,5,9) (DOK 1)
- two grams
- six grams
- one gram
- Which of the following compounds contains the largest number of atoms? (Outcome # 1,9) (DOK 2)
- 1.00 mole of NH3
- 1.00 mole of H2SO4
- 1.00 mole of HBr
- How many atoms are in 12.8 g of copper metal? (Outcome # 6) (DOK 2)
- 7.71 x 1024 atoms
- 1.21 x 1023 atoms
- 3.34 x 10-25 atoms
- How many grams contain 1.45 x 1021 zinc atoms? (Outcome # 6) (DOK 2)
- 1.57 x 10-1 g
- 5.71 x 1046 g
- 2.41 x 10-3 g
- How many moles are in 11.0 g of silver metal? (Outcome # 6) (DOK 1)
- 9.81 moles
- 1.19 x 103 moles
- 1.02 x 10-1 moles
- What is Avogadro’s number? (Outcome # 6) (DOK 1)
- 6.022 x 1023 particles/mole
- 6.022 x 1022 particles/mole
- 6.022 x 1024 particles/mole
- Which of the following 85.0 g samples contain the greatest number of atoms? (Outcome # 6) (DOK 3)
- zinc
- gold
- zirconium
- Which of the following 49.5 g samples contain the fewest number of atoms? (Outcome # 6) (DOK 3) (Paired item 1)
- lead
- platinum
- iron
- How many atoms are in 1 mole of calcium? (Outcome # 6) (DOK 1)
- 6.022 x 1023 atoms
- 6.022 x 10-23 atoms
- 6.022 x 1022 atoms
- How many moles contain 6.022 x 1023 lithium atoms? (Outcome # 6) (DOK 1)
- 6.022 x 1023 moles
- 1 mole
- 2 moles
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