{"type":"doc","content":[{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q1. What is the molecular geometry of PBr3?"}]},{"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: Molecular Geometry (VSEPR Theory)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how to determine the shape of a molecule based on the number of bonding and lone pairs around the central atom."}]},{"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":"VSEPR Theory: Valence Shell Electron Pair Repulsion theory predicts the geometry of molecules."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Lone pairs: Non-bonding pairs of electrons on the central atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Bonding pairs: Pairs of electrons shared between atoms."}]}]}]},{"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":"Draw the Lewis structure for PBr3. Identify the central atom (P) and count the number of valence electrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine the number of bonding pairs (P-Br bonds) and lone pairs on phosphorus."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use VSEPR theory to predict the arrangement of electron domains around phosphorus."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the molecular geometry based on the arrangement of atoms (ignore lone pairs for geometry)."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q2. Which of the following species does not have tetrahedral molecular geometry?"}]},{"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: Molecular Geometry (VSEPR Theory)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question asks you to recognize which species among the options does not have a tetrahedral shape."}]},{"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":"Tetrahedral geometry: Four electron domains arranged around a central atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"VSEPR Theory: Used to predict shapes based on electron domain count."}]}]}]},{"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":"For each species, draw the Lewis structure and count the number of electron domains around the central atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine if the arrangement of domains matches tetrahedral geometry (four domains, all bonding)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Check for lone pairs or double bonds that would alter the geometry."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which species does not fit the tetrahedral geometry."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q3. Which of the following species is polar?"}]},{"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: Molecular Polarity"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine whether a molecule is polar or nonpolar based on its shape and bond polarity."}]},{"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":"Polarity: A molecule is polar if it has an uneven distribution of charge."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Electronegativity: Difference in electronegativity between atoms can lead to polar bonds."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Molecular geometry: Shape affects whether dipoles cancel."}]}]}]},{"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":"For each species, determine the molecular geometry using VSEPR theory."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Assess the symmetry of the molecule and whether the dipole moments cancel."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which molecule has a net dipole moment (is polar)."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q4. Which of the following species is nonpolar?"}]},{"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: Molecular Polarity"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question asks you to identify a molecule that is nonpolar, based on its shape and bond polarity."}]},{"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":"Nonpolar molecule: No net dipole moment; often symmetrical."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"VSEPR Theory: Used to determine shape."}]}]}]},{"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":"Draw the Lewis structure for each species and determine its geometry."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Check if the molecule is symmetrical and if the dipole moments cancel."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify which molecule is nonpolar."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q5. How many pi bonds are present in CO2, O=C=O?"}]},{"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: Bond Types (Sigma and Pi Bonds)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the types of bonds present in a molecule, specifically pi bonds."}]},{"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":"Pi bond ("},{"type":"inlineMath","attrs":{"latex":"\\pi"}},{"type":"text","text":"): Formed by sideways overlap of p orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Sigma bond ("},{"type":"inlineMath","attrs":{"latex":"\\sigma"}},{"type":"text","text":"): Formed by head-on overlap of orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Double bond: Contains one sigma and one pi bond."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Triple bond: Contains one sigma and two pi bonds."}]}]}]},{"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":"Draw the Lewis structure for CO2 and identify the types of bonds between carbon and oxygen."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the number of double bonds and recall that each double bond has one pi bond."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up the total number of pi bonds present in the molecule."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q6. From left to right, give the hybridization of each carbon atom in H3C-C≡CH."}]},{"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: Hybridization"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to assign hybridization to each carbon atom in a molecule based on their bonding and geometry."}]},{"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":"Hybridization: Mixing of atomic orbitals to form new hybrid orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"sp3: Four electron domains (tetrahedral)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"sp2: Three electron domains (trigonal planar)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"sp: Two electron domains (linear)."}]}]}]},{"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":"Draw the structure of H3C-C≡CH and label each carbon atom from left to right."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the number of electron domains (bonds and lone pairs) around each carbon atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Assign the correct hybridization based on the domain count for each 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q7. How many orbitals does a set of sp2 hybrid orbitals contain?"}]},{"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: Hybridization"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your knowledge of how many hybrid orbitals are formed when an atom undergoes sp2 hybridization."}]},{"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":"sp2 hybridization: Combination of one s and two p orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Hybrid orbitals: New orbitals formed from mixing atomic orbitals."}]}]}]},{"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 sp2 hybridization involves mixing one s and two p orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the total number of hybrid orbitals formed (equal to the number of atomic orbitals mixed)."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q8. How many p atomic orbitals are required to generate a set of sp3 hybrid orbitals?"}]},{"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: Hybridization"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of the atomic orbitals involved in sp3 hybridization."}]},{"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":"sp3 hybridization: Combination of one s and three p orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Atomic orbitals: s and p orbitals from the same atom."}]}]}]},{"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 sp3 hybridization involves mixing one s and three p orbitals."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the number of p orbitals used in the hybridization process."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q9. Ammonia (NH3) is a trigonal pyramidal molecule with H-N-H bond angles of about 107°; describe the bonding using hybrid orbital theory."}]},{"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: Hybridization and Molecular Geometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question asks you to use hybrid orbital theory to explain the bonding and geometry of ammonia."}]},{"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":"Hybridization: Determines the shape and bond angles."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Trigonal pyramidal: Three bonds and one lone pair."}]}]}]},{"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":"Draw the Lewis structure for NH3 and count the electron domains around nitrogen."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine the hybridization based on the number of domains."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Relate the hybridization to the observed bond angles and geometry."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q10. Use hybrid orbital theory to describe the bonding and bond angles of about 84.8° in BrF5."}]},{"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: Hybridization and Molecular Geometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to use hybrid orbital theory to explain the bonding and bond angles in BrF5."}]},{"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":"Hybridization: Determines the arrangement of electron domains."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Bond angles: Influenced by lone pairs and geometry."}]}]}]},{"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":"Draw the Lewis structure for BrF5 and count the electron domains around bromine."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine the hybridization based on the number of domains."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Relate the hybridization to the observed bond angles and geometry."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q11. For the element fluorine (F), determine the number of unpaired electrons in the Lewis Dot Symbol."}]},{"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: Lewis Dot Structures and Electron Configuration"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how to represent valence electrons and identify unpaired electrons for an element."}]},{"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":"Lewis Dot Symbol: Shows valence electrons as dots around the element symbol."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Unpaired electrons: Electrons not paired in the valence shell."}]}]}]},{"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 electron configuration for fluorine and identify the valence electrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Draw the Lewis dot symbol and count the number of unpaired electrons."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q12. According to valence bond theory, how many bonds would you expect a nitrogen atom to form in its ground state?"}]},{"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: Valence Bond Theory"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how many bonds an atom can form based on its electron configuration."}]},{"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":"Valence bond theory: Explains bonding based on available unpaired electrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Ground state: The lowest energy electron configuration."}]}]}]},{"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 electron configuration for nitrogen."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the number of unpaired electrons in the valence shell."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Relate the number of unpaired electrons to the number of bonds nitrogen can 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q13. What is the most likely electron-domain geometry of the left carbon of H2C=CH2?"}]},{"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: Electron-Domain Geometry (VSEPR Theory)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine the electron-domain geometry based on the number of electron domains around a carbon atom."}]},{"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":"Electron-domain geometry: Arrangement of all electron domains (bonding and lone pairs)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"VSEPR Theory: Used to predict geometry."}]}]}]},{"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":"Draw the structure of H2C=CH2 and focus on the left carbon atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the number of electron domains (bonds and lone pairs) around the left carbon."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use VSEPR theory to determine the electron-domain geometry."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q14. What is the most likely molecular geometry around the left carbon of H2C=CH2?"}]},{"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: Molecular Geometry (VSEPR Theory)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine the molecular geometry based on the arrangement of atoms around the left carbon."}]},{"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":"Molecular geometry: Arrangement of atoms (ignoring lone pairs)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"VSEPR Theory: Used to predict geometry."}]}]}]},{"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":"Draw the structure of H2C=CH2 and focus on the left carbon atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the number of bonding domains around the left carbon."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use VSEPR theory to determine the molecular geometry."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q15. What is the most likely bond angle for the H-C-H of the left carbon of H2C=CH2?"}]},{"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: Bond Angles and Molecular Geometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of how geometry affects bond angles."}]},{"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":"Bond angle: The angle between two bonds originating from the same atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"VSEPR Theory: Used to predict bond angles based on geometry."}]}]}]},{"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":"Determine the electron-domain geometry of the left carbon."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the typical bond angle for that geometry."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Apply this to the H-C-H bond angle."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q16. Which of the following atoms must always obey the octet rule?"}]},{"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: Octet Rule"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your understanding of which elements strictly follow the octet rule when forming compounds."}]},{"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":"Octet rule: Atoms tend to have eight electrons in their valence shell."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Exceptions: Some elements can have expanded or incomplete octets."}]}]}]},{"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 which elements are known to always obey the octet rule."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Consider the position of each atom in the periodic table and their typical bonding behavior."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q17. How many lone pairs are on the central atom in XeOF2?"}]},{"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: Lewis Structures and Lone Pairs"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to draw a Lewis structure and count the lone pairs on the central atom."}]},{"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":"Lone pairs: Non-bonding pairs of electrons on the central atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Lewis structure: Shows bonding and lone pairs."}]}]}]},{"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":"Draw the Lewis structure for XeOF2."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Count the number of valence electrons for xenon and distribute them among bonds and lone pairs."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Identify the number of lone pairs remaining on xenon after bonding."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q18. Determine the formal charge on the carbon atom of carbon monoxide, :C≡O:."}]},{"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: Formal Charge Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate formal charge using the Lewis structure."}]},{"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":"Formal charge formula: "},{"type":"inlineMath","attrs":{"latex":"\\text{Formal charge} = \\text{Valence electrons} - (\\text{Nonbonding electrons} + \\frac{1}{2}\\text{Bonding electrons})"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Lewis structure: Shows bonding and lone pairs."}]}]}]},{"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":"Draw the Lewis structure for CO and identify the number of bonds and lone pairs on carbon."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Apply the formal charge formula to the carbon atom."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Calculate the values for each term in the 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q19. Classify the following bond, O=O."}]},{"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: Bond Types (Polarity and Ionic/Covalent)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to classify a bond as nonpolar, polar, or ionic."}]},{"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":"Nonpolar covalent bond: Equal sharing of electrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Polar covalent bond: Unequal sharing of electrons."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Ionic bond: Transfer of electrons."}]}]}]},{"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 elements involved and their electronegativities."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine if the bond is between identical atoms (nonpolar) or different atoms (polar/ionic)."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q20. Classify the following bond, CsBr."}]},{"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: Bond Types (Polarity and Ionic/Covalent)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to classify a bond as nonpolar, polar, or ionic."}]},{"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":"Ionic bond: Formed between metals and nonmetals with large electronegativity differences."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Polar covalent bond: Moderate difference in electronegativity."}]}]}]},{"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 elements involved and their positions in the periodic table."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine the type of bond based on electronegativity difference."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q21. Classify the following bond, Cl-F."}]},{"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: Bond Types (Polarity and Ionic/Covalent)"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to classify a bond as nonpolar, polar, or ionic."}]},{"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":"Polar covalent bond: Unequal sharing of electrons between atoms with different electronegativities."}]}]}]},{"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 elements involved and their electronegativities."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Determine if the bond is polar based on the difference in electronegativity."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q22. What is the correct name for CaSO4?"}]},{"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: Nomenclature of Ionic Compounds"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to name ionic compounds using standard nomenclature rules."}]},{"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":"Ionic compound: Made of a metal and a polyatomic ion."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Polyatomic ions: SO42- is sulfate."}]}]}]},{"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 cation (Ca2+) and the anion (SO42-)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Use the correct name for the polyatomic ion."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Combine the names to form the compound name."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q23. What is the correct formula for nickel(II) perchlorate?"}]},{"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: Nomenclature and Formula Writing"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to write the correct formula for a compound given its name."}]},{"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":"Nickel(II): Ni2+"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Perchlorate: ClO4-"}]}]}]},{"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 charges of the cation and anion."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Balance the charges to write the correct 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q24. Determine the molar mass of Co(NO2)2."}]},{"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 its constituent 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: Sum of atomic masses in one mole of a compound."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Atomic masses: Co, N, O (use periodic table values)."}]}]}]},{"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 formula and count the number of each atom in Co(NO2)2."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the atomic mass of each element by the number of atoms present."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up the total mass to get the 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q25. Determine the molar mass of Fe2(SO4)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 its constituent 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: Sum of atomic masses in one mole of a compound."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Atomic masses: Fe, S, O (use periodic table values)."}]}]}]},{"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 formula and count the number of each atom in Fe2(SO4)3."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the atomic mass of each element by the number of atoms present."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up the total mass to get the 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q26. How many moles of hydrogen atoms are there in 6.50 g of ammonia (NH3)?"}]},{"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 Calculations and Stoichiometry"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert mass to moles and account for the number of hydrogen atoms in a compound."}]},{"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 NH3: "},{"type":"inlineMath","attrs":{"latex":"\\text{N} = 14.01"}},{"type":"text","text":", "},{"type":"inlineMath","attrs":{"latex":"\\text{H} = 1.008"}}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Number of hydrogen atoms per molecule: 3"}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Moles} = \\frac{\\text{mass}}{\\text{molar 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":"Calculate the molar mass of NH3."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert 6.50 g NH3 to moles using the molar mass."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of moles of NH3 by the number of hydrogen atoms per molecule."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q27. Determine the empirical formula of a compound that has the following composition: 92.3% C and 7.7% H."}]},{"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 Formula Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine the simplest whole-number ratio of atoms in a compound based on percent composition."}]},{"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 ratio of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Percent composition: Used to find moles of each element."}]}]}]},{"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 100 g of the compound, so you have 92.3 g C and 7.7 g H."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert grams to moles for each element using their atomic masses."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide each by the smallest number of moles to get the ratio."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q28. Determine the empirical formula of a compound that is: 48.6% C, 8.2% H, and 43.2% O."}]},{"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 Formula Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to determine the simplest whole-number ratio of atoms in a compound based on percent composition."}]},{"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 ratio of atoms."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Percent composition: Used to find moles of each element."}]}]}]},{"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 100 g of the compound, so you have 48.6 g C, 8.2 g H, and 43.2 g O."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Convert grams to moles for each element using their atomic masses."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide each by the smallest number of moles to get the ratio."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q29. Determine the mass in grams of 2.75 moles of glucose (C6H12O6)."}]},{"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 Conversion"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert moles to mass using the molar mass of a compound."}]},{"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 atomic masses for C6H12O6."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Mass} = \\text{moles} \\times \\text{molar 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":"Calculate the molar mass of glucose (C6H12O6)."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the number of moles by the molar mass to get the mass in grams."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q30. Determine the number of moles in 59.8 g of sodium nitrate (NaNO3)."}]},{"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 Conversion"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to convert mass to moles using the molar mass of a compound."}]},{"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 atomic masses for NaNO3."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"inlineMath","attrs":{"latex":"\\text{Moles} = \\frac{\\text{mass}}{\\text{molar 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":"Calculate the molar mass of NaNO3."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Divide the mass (59.8 g) by the molar mass to get the number of moles."}]}]}]},{"type":"heading","attrs":{"textAlign":null,"level":4},"content":[{"type":"text","marks":[{"type":"underline"}],"text":"Try solving on your own before revealing the answer!"}]},{"type":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q31. Determine the molecular mass of citric acid, H3C6H5O7."}]},{"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: Molecular Mass Calculation"}]},{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"This question tests your ability to calculate the molecular mass (in amu) of a compound by summing the atomic masses of its constituent 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":"Molecular mass: Sum of atomic masses for all atoms in the formula."}]}]}]},{"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":"Count the number of each atom in H3C6H5O7."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Multiply the atomic mass of each element by the number of atoms present."}]}]},{"type":"listItem","content":[{"type":"paragraph","attrs":{"textAlign":null},"content":[{"type":"text","text":"Add up the total mass to get the molecular 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":"heading","attrs":{"textAlign":null,"level":3},"content":[{"type":"text","marks":[{"type":"bold"}],"text":"Q32. Determine the percent composition of carbon by mass in acetaminophen, C8H9NO2."}]},{"type":"heading","attrs":{"textAlign":null,"