General Chemistry Exam 2 Study Guidance

{"type":"doc","content":[{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q1. An element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The heavier two isotopes have an abundance of 4.68% and 3.09%, respectively. What is the mass of the element?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Average Atomic Mass"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the average atomic mass of an element using the masses and percent abundances of its isotopes."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Isotope: Atoms of the same element with different numbers of neutrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Percent abundance: The percentage of each isotope in a natural sample."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Average atomic mass} = \\sum (\\text{isotope mass} \\times \\text{fractional abundance})"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert the percent abundances of the heavier two isotopes to decimal form (divide by 100)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the percent abundance of the lightest isotope by subtracting the sum of the other two from 100%."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert the abundance of the lightest isotope to decimal form."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the calculation for average atomic mass using the formula above, multiplying each isotope's mass by its fractional abundance."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 28.085 amu"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We calculated the weighted average using the given abundances and masses."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q2. How many argon atoms are contained in "},{"type":"inlineMath","attrs":{"latex":"7.66 \\times 10^5"}},{"type":"text","marks":[{"type":"bold"}],"text":" mmol of argon?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Mole Concept and Avogadro's Number"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert millimoles to moles and then to number of atoms using Avogadro's number."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Millimole (mmol): "},{"type":"inlineMath","attrs":{"latex":"1~\\text{mmol} = 10^{-3}~\\text{mol}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Avogadro's number: "},{"type":"inlineMath","attrs":{"latex":"6.022 \\times 10^{23}"}},{"type":"text","text":" atoms/mol"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Number of atoms} = \\text{moles} \\times 6.022 \\times 10^{23}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert "},{"type":"inlineMath","attrs":{"latex":"7.66 \\times 10^5"}},{"type":"text","text":" mmol to moles by dividing by 1000."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of moles by Avogadro's number to get the number of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the calculation, but do not compute the final value yet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: "},{"type":"inlineMath","attrs":{"latex":"4.61 \\times 10^{26}"}},{"type":"text","marks":[{"type":"bold"}],"text":" Ar atoms"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We converted mmol to mol and then multiplied by Avogadro's number."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q3. What mass (in g) does 7.98 moles of Kr have?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Moles and Molar Mass"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use the molar mass of krypton to convert moles to grams."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Mole: The SI unit for amount of substance."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass: Mass of one mole of a substance (g/mol)."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{mass (g)} = \\text{moles} \\times \\text{molar mass (g/mol)}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Look up the molar mass of Kr (krypton) on the periodic table."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of moles (7.98) by the molar mass of Kr."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the calculation, but do not compute the final value yet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 668 g"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We used the molar mass of Kr (about 83.8 g/mol) and multiplied by the number of moles."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q4. How many Zn atoms are contained in 922 g of Zn?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Mass-Mole-Atom Conversions"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert mass to moles and then to atoms using Avogadro's number."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass of Zn: about 65.38 g/mol"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Avogadro's number: "},{"type":"inlineMath","attrs":{"latex":"6.022 \\times 10^{23}"}},{"type":"text","text":" atoms/mol"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formulas:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles} = \\frac{\\text{mass (g)}}{\\text{molar mass (g/mol)}}"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{atoms} = \\text{moles} \\times 6.022 \\times 10^{23}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the number of moles of Zn by dividing 922 g by the molar mass of Zn."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of moles by Avogadro's number to get the number of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the calculation, but do not compute the final value yet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: "},{"type":"inlineMath","attrs":{"latex":"8.49 \\times 10^{24}"}},{"type":"text","marks":[{"type":"bold"}],"text":" Zn atoms"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We converted grams to moles and then moles to atoms using Avogadro's number."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q5. Calculate the molar mass of Al(C2H3O2)3."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Molar Mass Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the molar mass of a compound by summing the atomic masses of all atoms in the formula."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass: Sum of the atomic masses of all atoms in a formula."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Atomic masses: Al (26.98), C (12.01), H (1.008), O (16.00)"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Molar mass} = \\sum (\\text{number of atoms} \\times \\text{atomic mass})"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Break down the formula: Al(C2H3O2)3 means 1 Al, 6 C, 9 H, and 6 O."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of each atom by its atomic mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up all the contributions to get the total molar mass."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 204.13 g/mol"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We summed the atomic masses for all atoms in the formula."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q6. Calculate the molar mass of H2CO3."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Molar Mass Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to sum atomic masses for a molecular formula."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Atomic masses: H (1.008), C (12.01), O (16.00)"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Molar mass} = 2 \\times \\text{H} + 1 \\times \\text{C} + 3 \\times \\text{O}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of each atom by its atomic mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up all the contributions to get the total molar mass."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 62.03 g/mol"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We summed the atomic masses for all atoms in the formula."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q7. How many atoms of carbon are contained in 47.6 g of Al2(CO3)3? The molar mass of Al2(CO3)3 is 233.99 g/mol."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Mass-Mole-Atom Conversions"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert mass to moles, then moles of compound to moles of carbon, and finally to atoms."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass: 233.99 g/mol"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Avogadro's number: "},{"type":"inlineMath","attrs":{"latex":"6.022 \\times 10^{23}"}},{"type":"text","text":" atoms/mol"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formulas:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of compound} = \\frac{\\text{mass}}{\\text{molar mass}}"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of C} = \\text{moles of compound} \\times 3"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{atoms of C} = \\text{moles of C} \\times 6.022 \\times 10^{23}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of Al2(CO3)3 by dividing 47.6 g by 233.99 g/mol."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the moles of compound by 3 to get moles of carbon (since each formula unit has 3 C atoms)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply moles of carbon by Avogadro's number to get atoms of carbon."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: "},{"type":"inlineMath","attrs":{"latex":"3.68 \\times 10^{23}"}},{"type":"text","marks":[{"type":"bold"}],"text":" C atoms"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We converted mass to moles, then to moles of C, then to atoms."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q8. Determine the molecular formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Empirical and Molecular Formula Determination"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert percent composition to a molecular formula."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Empirical formula: Simplest whole-number ratio of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molecular formula: Actual number of atoms in a molecule."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key steps:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"1. Assume 100 g sample, convert % to grams."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"2. Convert grams to moles for each element."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"3. Find the simplest whole-number ratio."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Assume a 100 g sample: use the percentages as grams."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert grams of each element to moles using their atomic masses."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide each mole value by the smallest number of moles to get the ratio."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply to get whole numbers if needed, then write the empirical formula."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: C18H27NO3"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We determined the empirical formula and matched it to the molecular formula."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q9. Which one of the following is NOT an empirical formula?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Empirical vs. Molecular Formulas"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to identify empirical formulas, which are the simplest whole-number ratios of atoms."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Empirical formula: Simplest whole-number ratio of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molecular formula: Actual number of atoms in a molecule, may not be simplest ratio."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Review each formula and check if the subscripts can be reduced to a simpler ratio."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which formula is not in its simplest form."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: C2H6O2"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This formula can be reduced to CH3O, so it is not empirical."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q10. Determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. The molecular weight is 194.19 g/mol."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Empirical and Molecular Formula Determination"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine the empirical formula from percent composition and then use the molecular weight to find the molecular formula."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Empirical formula: Simplest whole-number ratio of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molecular formula: Actual number of atoms in a molecule."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key steps:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"1. Assume 100 g sample, convert % to grams."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"2. Convert grams to moles for each element."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"3. Find the simplest whole-number ratio."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"4. Calculate empirical formula mass and compare to molecular weight."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Assume a 100 g sample: use the percentages as grams."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert grams of each element to moles using their atomic masses."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide each mole value by the smallest number of moles to get the ratio."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the empirical formula mass and determine how many times it fits into the molecular weight."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: C8H10N4O2"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We determined the empirical formula and scaled it to match the molecular weight."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q11. What is the percent yield of H2 if 58.94g of methane yields 14.61g of H2 for the following reaction: CH4(g) + 2 H2O(g) → 4 H2(g) + CO2(g)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Percent Yield Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate percent yield from actual and theoretical yields."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Percent yield: "},{"type":"inlineMath","attrs":{"latex":"\\frac{\\text{actual yield}}{\\text{theoretical yield}} \\times 100"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Theoretical yield: Maximum amount of product possible from limiting reactant."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of CH4 from its mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use stoichiometry to find theoretical moles of H2 produced."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert theoretical moles of H2 to grams."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the percent yield formula using actual and theoretical yields."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 49.21%"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We calculated the theoretical yield and compared it to the actual yield."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q12. How many moles of nitrogen are formed when 58.6 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11 g/mol. 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Stoichiometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use mass-mole conversions and reaction stoichiometry."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass: 101.11 g/mol"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Stoichiometric coefficients: 4 KNO3 → 2 N2"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formulas:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of KNO3} = \\frac{58.6}{101.11}"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of N2} = \\text{moles of KNO3} \\times \\frac{2}{4}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of KNO3 by dividing mass by molar mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the stoichiometric ratio to find moles of N2 produced."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the calculation, but do not compute the final value yet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 0.290 mol N2"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We used stoichiometry to convert moles of KNO3 to moles of N2."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q13. Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N2O4 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol. N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Limiting Reactant and Stoichiometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to identify the limiting reactant and calculate the mass of product formed."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Limiting reactant: The reactant that is completely consumed first."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Stoichiometry: Use coefficients to relate reactants and products."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key steps:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"1. Convert mass of each reactant to moles."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"2. Use stoichiometry to determine which reactant limits the reaction."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"3. Use limiting reactant to calculate moles of N2 produced."}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"4. Convert moles of N2 to grams."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of N2O4 and N2H4 using their masses and molar masses."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the reaction coefficients to determine which reactant will run out first."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the limiting reactant to calculate moles of N2 produced."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert moles of N2 to grams using its molar mass."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: LR = N2H4, 59.0 g N2 formed"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We identified the limiting reactant and calculated the mass of N2 produced."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q14. Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Percent Yield Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate percent yield from actual and theoretical yields."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Percent yield: "},{"type":"inlineMath","attrs":{"latex":"\\frac{\\text{actual yield}}{\\text{theoretical yield}} \\times 100"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Theoretical yield: Maximum amount of product possible from limiting reactant."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of Fe2O3 from its mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use stoichiometry to find theoretical moles of Fe produced."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert theoretical moles of Fe to grams."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the percent yield formula using actual and theoretical yields."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 61.03 %"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We calculated the theoretical yield and compared it to the actual yield."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q15. Identify acetic acid."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electrolytes and Acid Strength"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to classify acetic acid as a strong/weak acid and strong/weak electrolyte."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Strong acid: Completely ionizes in water."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Weak acid: Partially ionizes in water."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Strong electrolyte: Conducts electricity well; fully dissociates."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Weak electrolyte: Conducts electricity poorly; partially dissociates."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Recall that acetic acid (CH3COOH) is a weak acid."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine if it is a strong or weak electrolyte based on its ionization in water."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: weak electrolyte, weak acid"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Acetic acid only partially ionizes in water."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q16. Identify HCl."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Electrolytes and Acid Strength"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to classify HCl as a strong/weak acid and strong/weak electrolyte."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Strong acid: Completely ionizes in water."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Strong electrolyte: Conducts electricity well; fully dissociates."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Recall that HCl is a strong acid."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine if it is a strong or weak electrolyte based on its ionization in water."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: strong electrolyte, strong acid"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"HCl fully ionizes in water and conducts electricity well."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q17. Which one of the following compounds is soluble in water?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Solubility Rules"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to apply solubility rules to determine which compound is soluble in water."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solubility rules: Provided in the formula sheet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Review each compound and apply the solubility rules from the formula sheet."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which compound is always soluble according to the rules."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: Ni(NO3)2"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Nitrate salts are always soluble."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q18. Which of the following compounds is soluble in water?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Solubility Rules"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to apply solubility rules to determine which compound is soluble in water."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solubility rules: Provided in the formula sheet."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Review each compound and apply the solubility rules from the formula sheet."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which compound is always soluble according to the rules."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: CaS"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"CaS is soluble; the others are not according to the rules."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q19. Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 mL of solution."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Molarity Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate molarity from mass of solute and volume of solution."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molarity (M): "},{"type":"inlineMath","attrs":{"latex":"\\text{mol solute} / \\text{L solution}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molar mass of LiBr: about 86.85 g/mol"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"M = \\frac{\\text{moles of solute}}{\\text{liters of solution}}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of LiBr by dividing mass by molar mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert 750.0 mL to liters."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide moles of LiBr by liters of solution to get molarity."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 1.50 M"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We calculated moles and divided by volume in liters."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q20. According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that there is excess FeCl2. FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Solution Stoichiometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use molarity and volume to find moles of reactant and then use stoichiometry to find moles of product."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molarity (M): "},{"type":"inlineMath","attrs":{"latex":"\\text{mol/L}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Stoichiometry: Use coefficients to relate reactants and products."}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formulas:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of LiOH} = M \\times V"}}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{moles of Fe(OH)2} = \\frac{\\text{moles of LiOH}}{2}"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert 175.0 mL to liters."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate moles of LiOH using molarity and volume."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use stoichiometry to find moles of Fe(OH)2 produced."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: "},{"type":"inlineMath","attrs":{"latex":"1.99 \\times 10^{-2}"}},{"type":"text","marks":[{"type":"bold"}],"text":" moles"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We used molarity, volume, and stoichiometry to find moles of product."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q21. Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of K2S and Fe(NO3)2 are mixed."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Net Ionic Equations"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to write net ionic equations for precipitation reactions."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Net ionic equation: Shows only the species that actually change during the reaction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solubility rules: Used to determine if a precipitate forms."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Write the balanced molecular equation for the reaction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the ions present in solution."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Apply solubility rules to determine if a precipitate forms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Write the net ionic equation, showing only the formation of the precipitate."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: Fe2+(aq) + S2-(aq) → FeS(s)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Only the ions forming the precipitate are shown in the net ionic equation."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q22. Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of lithium sulfide and copper (II) nitrate are mixed."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Complete Ionic Equations"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to write complete ionic equations for precipitation reactions."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Complete ionic equation: Shows all ions present in solution."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Solubility rules: Used to determine if a precipitate forms."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Write the balanced molecular equation for the reaction."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Break all soluble compounds into their ions."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Show the formation of any precipitate as a solid."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 2 Li+(aq) + S2-(aq) + Cu2+(aq) + 2 NO3-(aq) → CuS(s) + 2 Li+(aq) + 2 NO3-(aq)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"All soluble ions are shown, and the precipitate is written as a solid."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q23. How many milliliters of a 9.0 M H2SO4 solution are needed to make 0.45 L of a 3.5 M solution?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Solution Dilution"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use the dilution equation to calculate volumes."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molarity (M): "},{"type":"inlineMath","attrs":{"latex":"\\text{mol/L}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Volume (V): in liters or milliliters"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"M_1 V_1 = M_2 V_2"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the initial and final concentrations and volumes."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the dilution equation to solve for the unknown volume."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert the final answer to milliliters if needed."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"collapsible","content":[{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Final Answer: 0.18 L (or 180 mL)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"We used the dilution equation to solve for the required volume."}]}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q24. What is the concentration of HCl in the final solution when 65 mL of a 9.0 M HCl solution is diluted with pure water to a total volume of 0.15 L?"}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Background"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Topic: Solution Dilution"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use the dilution equation to calculate the final concentration."}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Key Terms and Formulas"}]},{"type":"bulletList","content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molarity (M): "},{"type":"inlineMath","attrs":{"latex":"\\text{mol/L}"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Volume (V): in liters or milliliters"}]}]}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Key formula:"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"M_1 V_1 = M_2 V_2"}}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step-by-Step Guidance"}]},{"type":"orderedList","attrs":{"start":1,"type":null},"content":[{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the initial concentration and volume, and the final volume."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Set up the dilution equation to solve for the final concentration."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert volumes to the same units before calculating."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your o