Essential knowledge statements from the AP Chemistry CED: - One cannot count particles directly while performing laboratory connection between the masses of substances reacting and the ac undergoing chemical changes. Avogadro's number (N_(A)=6.02times 10^23mol^-1) provides the connect in a pure sample of a substance and the number of constituent par substance. - Expressing the mass of an individual atom or molecule in atomic because the average mass in amu of one particle (atom or molecul will always be numerically equal to the molar mass of that substan quantitative connection between the mass of a substance and the n substance contains. __ 1. Calculate the mass in grams, of 0.0850 mol Ba(OH)_(2) 2. Calculate the number of moles of C_(4)H_(10) present in 2.00 g C_(4)H_(10) 3. Calculate the number of atoms of Si present in 35.0 mol Si. . Calculate the number of moles of O_(3) present in 4.3times 10^24 molecules

1. To calculate the mass in grams of 0.0850 mol $Ba(OH)_{2}$, we need to use the molar mass of $Ba(OH)_{2}$. The molar mass of $Ba(OH)_{2}$ is 171.34 g/mol. Therefore, the mass in grams of 0.0850 mol $Ba(OH)_{2}$ is:<br /><br />Mass = Moles × Molar Mass<br />Mass = 0.0850 mol × 171.34 g/mol<br />Mass = 14.63 g<br /><br />2. To calculate the number of moles of $C_{4}H_{10}$ present in 2.00 g $C_{4}H_{10}$, we need to use the molar mass of $C_{4}H_{10}$. The molar mass of $C_{4}H_{10}$ is 58.12 g/mol. Therefore, the number of moles of $C_{4}H_{10}$ present in 2.00 g $C_{4}H_{10}$ is:<br /><br />Moles = Mass / Molar Mass<br />Moles = 2.00 g / 58.12 g/mol<br />Moles = 0.0345 mol<br /><br />3. To calculate the number of atoms of Si present in 350 mol Si, we need to use Avogadro's number ($N_{A}=6.02\times 10^{23}mol^{-1}$). Therefore, the number of atoms of Si present in 350 mol Si is:<br /><br />Number of atoms = Moles × Avogadro's number<br />Number of atoms = 350 mol × 6.02 × 10^{23} mol^{-1}<br />Number of atoms = 2.107 × 10^{26} atoms<br /><br />4. To calculate the number of moles of $O_{3}$ present in $4.3\times 10^{24}$ molecules, we need to use Avogadro's number ($N_{A}=6.02\times 10^{23}mol^{-1}$). Therefore, the number of moles of $O_{3}$ present in $4.3\times 10^{24}$ molecules is:<br /><br />Number of moles = Number of molecules / Avogadro's number<br />Number of moles = $4.3\times 10^{24}$ molecules / 6.02 × 10^{23} mol^{-1}<br />Number of moles = 7.15 mol