This question tests the skill of moles and molar mass. Molar mass relates mass to moles, but this problem involves converting from particles to moles using Avogadro's number. Avogadro's number bridges moles to the number of atoms or molecules, allowing us to find moles by dividing the number of particles by 6.022 × $10^23$ per mole. For elemental aluminum, this calculation reveals the amount in moles from a count of atoms. A common distractor is choice C, 6.02 × $10^23$ mol, which comes from mistakenly multiplying instead of dividing by Avogadro's number, inverting the conversion factor. Ensure you divide when going from particles to moles. A transferable strategy is to write the unit you want (moles) and use Avogadro's number as the bridge by dividing particles by particles per mole.

This question tests the skill of moles and molar mass. Molar mass links mass to amount through moles = mass / molar mass, with KBr's molar mass as 39.1 + 79.9 = 119 g/mol. This allows direct calculation from given mass. Avogadro's number, 6.02 × 10²³, links moles to particles for further conversions if needed. A tempting distractor is choice A, 0.0500 mol, from a common unit-conversion error of using half the molar mass, perhaps treating Br as 40 instead of 79.9. A transferable strategy is to write the unit you want and use molar mass as the bridge.

This question tests the skill of moles and molar mass. Molar mass links mass to amount as moles = mass / molar mass, with ethanol at 46.0 g/mol giving 0.500 mol. This conversion is fundamental. Avogadro's number, 6.02 × 10²³, links moles to particles. A tempting distractor is choice A, 0.250 mol, from a common unit-conversion error of halving the mass. A transferable strategy is to write the unit you want and use molar mass as the bridge.

This question tests the skill of moles and molar mass. Molar mass provides the conversion factor between mass and moles through the relationship: moles = mass ÷ molar mass. With 44.0 g of CO₂ and a molar mass of 44.0 g/mol, we calculate: 44.0 g ÷ 44.0 g/mol = 1.00 mol. Avogadro's number would be necessary to find the number of molecules, but here we only need moles. A tempting wrong answer (choice A, 0.0227 mol) results from incorrectly inverting the calculation and dividing 1 by 44.0 instead of dividing 44.0 by 44.0. To solve mole conversions accurately, set up the calculation so that grams cancel out, leaving moles as your final unit.

This question tests the skill of moles and molar mass. While molar mass links mass to moles, Avogadro's number (6.02×10²³ particles/mol) connects moles to the number of particles (atoms, molecules, or formula units). To find atoms from moles, we multiply: 0.500 mol × 6.02×10²³ atoms/mol = 3.01×10²³ atoms. The molar mass of Mg would only be needed if we were converting between mass and moles, but here we're converting between moles and atoms. A common error (choice C, 6.02×10²³ atoms) results from forgetting to multiply by the number of moles and just using Avogadro's number directly. To convert from moles to particles, always multiply the number of moles by Avogadro's number to get the total particle count.

This problem tests moles and molar mass by converting mass to moles for a complex molecule. The molar mass of glucose (C₆H₁₂O₆) equals 6 carbon atoms (6 × 12.0 = 72.0 g/mol) plus 12 hydrogen atoms (12 × 1.0 = 12.0 g/mol) plus 6 oxygen atoms (6 × 16.0 = 96.0 g/mol), totaling 180.0 g/mol. To find moles: 18.0 g ÷ 180.0 g/mol = 0.100 mol. A common mistake would be to use 18.0 as the molar mass directly, giving 1.00 mol. The key strategy is to systematically calculate molar mass by counting each type of atom and multiplying by its atomic mass before adding them all together.

This problem tests the skill of moles and molar mass. Molar mass serves as the conversion factor between mass (in grams) and amount (in moles), allowing us to determine how many moles are in a given mass of substance. To convert from grams to moles, we divide the mass by the molar mass: moles = mass ÷ molar mass. For this water sample: moles = 18.0 g ÷ 18.0 g/mol = 1.00 mol. A common error would be multiplying instead of dividing (18.0 × 18.0 = 324), which doesn't appear here but would give an incorrect large value. When converting mass to moles, always divide by molar mass—think of it as 'how many molar mass units fit into your sample mass.'

This problem requires using moles and molar mass to convert from moles to mass. The molar mass of NH₃ is 17.0 g/mol, which tells us the mass of one mole. To find the mass of 0.500 mol, we multiply: 0.500 mol × 17.0 g/mol = 8.50 g. A student might mistakenly divide the molar mass by moles (17.0 ÷ 0.500 = 34.0), which gives twice the molar mass rather than half. The reliable method is dimensional analysis: start with moles, multiply by molar mass (g/mol), and verify that mol cancels to leave grams.

This problem involves moles and molar mass concepts. Avogadro's number bridges between moles and particles, with one mole containing 6.02×10²³ formula units regardless of the substance. To find the number of formula units from moles, we multiply: formula units = moles × NA. For this MgCl₂ sample: formula units = 0.100 mol × 6.02×10²³ mol⁻¹ = 6.02×10²² formula units. A tempting error would be forgetting to account for the 0.100 coefficient and selecting 6.02×10²³ (choice B), which represents one full mole. When converting moles to particles, multiply by Avogadro's number and carefully track your decimal places or powers of ten.

This problem tests converting moles to mass using moles and molar mass. The molar mass of aluminum (27.0 g/mol) tells us that one mole of aluminum has a mass of 27.0 grams. To find the mass of 0.100 mol, we multiply: 0.100 mol × 27.0 g/mol = 2.70 g. A typical mistake would be dividing (0.100 ÷ 27.0 = 0.00370), which gives the wrong value and units (mol²/g instead of g). Always check your units: when multiplying moles by g/mol, the mol units cancel, leaving grams as desired.