Mole-atom (ExtraExercises) Given: \( { }^{27} \mathrm{Al},{ }^{56} \mathrm{Fe},{ }^{12} \mathrm{C},{ }^{14} \mathrm{~N},{ }^{108} \mathrm{Ag},{ }^{207} \mathrm{~Pb},{ }^{20} \mathrm{Ne},{ }^{197} \mathrm{Au},{ }^{28} \mathrm{Si} \), \( N A=6 \times 10^{23} \) atoms \( / \) nole . 1) How many mole-atoms are there in 42 gof Aluminum? 2) How manywhatoms are there in \( 7.5 \times 10^{21} \) aloms of iron? 3) How many atoms are the re in 1.5 moles of carbon? 4) What is the mass of 0.65 moles of copperatoms? 5) What is the mass of \( 1.25 \times 10^{24} \) alous of nitrogen? 6) How many atoms are there in 18 g of silver? 7) Determine the nass of \( 6.25 \times 10^{20} \) atoms of lead? 8) How many atoms are there in 0.0135 nole of Neons 9) What is the nass of 0.0085 moles of gold? 10) Find the rbof moles of \( 2 g \) of silicon. v) Knowing that the nass of 0.0282 noles of chlorine is 19 , find its molar mass. 12) A silver necklace has a majsy 2 nog. Calculate the mbof moles contained in this rectelice, as uell as the nb of atoons.

Let's solve each of the problems step by step, extracting the known conditions and applying the necessary calculations. ### 1) How many mole-atoms are there in 42 g of Aluminum? **Known Conditions:** - Molar mass of Aluminum (\( \text{Al} \)) = 27 g/mol - Mass = 42 g **Calculation:** To find the number of moles, we use the formula: \[ n = \frac{m}{M} \] where \( n \) is the number of moles, \( m \) is the mass, and \( M \) is the molar mass. \[ n = \frac{42 \, \text{g}}{27 \, \text{g/mol}} \approx 1.56 \, \text{moles} \] ### 2) How many atoms are there in \( 7.5 \times 10^{21} \) atoms of iron? **Known Conditions:** - Number of atoms = \( 7.5 \times 10^{21} \) - Avogadro's number (\( N_A \)) = \( 6 \times 10^{23} \) atoms/mol **Calculation:** To find the number of moles, we use: \[ n = \frac{\text{Number of atoms}}{N_A} \] \[ n = \frac{7.5 \times 10^{21}}{6 \times 10^{23}} \approx 0.0125 \, \text{moles} \] ### 3) How many atoms are there in 1.5 moles of carbon? **Known Conditions:** - Number of moles = 1.5 moles - Avogadro's number (\( N_A \)) = \( 6 \times 10^{23} \) atoms/mol **Calculation:** \[ \text{Number of atoms} = n \times N_A \] \[ \text{Number of atoms} = 1.5 \times 6 \times 10^{23} = 9 \times 10^{23} \, \text{atoms} \] ### 4) What is the mass of 0.65 moles of copper atoms? **Known Conditions:** - Molar mass of Copper (\( \text{Cu} \)) = 63.5 g/mol - Number of moles = 0.65 moles **Calculation:** \[ m = n \times M \] \[ m = 0.65 \times 63.5 \approx 41.28 \, \text{g} \] ### 5) What is the mass of \( 1.25 \times 10^{24} \) atoms of nitrogen? **Known Conditions:** - Number of atoms = \( 1.25 \times 10^{24} \) - Molar mass of Nitrogen (\( \text{N} \)) = 14 g/mol **Calculation:** First, find the number of moles: \[ n = \frac{1.25 \times 10^{24}}{6 \times 10^{23}} \approx 2.08 \, \text{moles} \] Then, calculate the mass: \[ m = n \times M \] \[ m = 2.08 \times 14 \approx 29.12 \, \text{g} \] ### 6) How many atoms are there in 18 g of silver? **Known Conditions:** - Molar mass of Silver (\( \text{Ag} \)) = 107.87 g/mol - Mass = 18 g **Calculation:** First, find the number of moles: \[ n = \frac{18}{107.87} \approx 0.166 \, \text{moles} \] Then, calculate the number of atoms: \[ \text{Number of atoms} = n \times N_A \] \[ \text{Number of atoms} = 0.166 \times 6 \times 10^{23} \approx 1.00 \times 10^{23} \, \text{atoms} \] ### 7) Determine the mass of \( 6.25 \times 10^{20} \) atoms of lead. **Known Conditions:** - Number of atoms = \( 6.25 \times 10^{20} \) - Molar mass of Lead (\( \text{Pb} \)) = 207.2 g/mol **Calculation:** First, find the number of moles: \[ n = \frac{6.25 \times 10^{20}}{6 \times 10^{23}} \approx 1.04 \times 10^{-3} \, \text{moles} \] Then, calculate the mass: \[ m = n \times M \] \[ m = 1.04 \times 10^{-3} \times 207.2 \approx 0.216 \, \text{g} \] ### 8) How many atoms are there in 0.0135 moles of Neon? **Known Conditions:** - Number of moles = 0.0135 moles - Avogadro's number (\( N_A \)) = \( 6 \times 10^{23} \) atoms/mol **Calculation:** \[ \text{Number of atoms} = n \times N_A \] \[ \text{Number of atoms} = 0.0135 \times 6 \times 10^{23} \approx 8.1 \times 10^{21} \, \text{atoms} \] ### 9) What is the mass of 0.0085 moles of gold? **Known Conditions:** - Molar mass of Gold (\( \text{Au} \)) = 197 g/mol - Number of moles = 0.0085 moles **Calculation:** \[ m = n \times M \] \[ m = 0.0085 \times 197 \approx 1.67 \, \text{g} \] ### 10) Find the number of moles of 2 g of silicon. **Known Conditions:** - Molar mass of Silicon (\( \text{Si} \)) = 28.09 g/mol - Mass = 2 g **Calculation:** \[ n = \frac{m}{M} \] \[ n = \frac{2}{28.09} \approx 0.071 \, \text{moles} \] ### 11) Knowing that the mass of 0.0282 moles of chlorine is 19 g, find its molar mass. **Known Conditions:** - Number of moles = 0.0282 moles - Mass = 19 g **Calculation:** \[ M = \frac{m}{n} \