NSTRUCTIONS Write the procedures and the answers to each exercise. If you he result (without the procedure) your answer will not be valid. Mark your final answers with a different color. Round your answers to two decimal places. .Convert 400 g of Mg(CN)_(2) into mol. Calculate the number of formula units in 600 g of NaNO_(2) . Convert 8.5times 10^27 molecules of KMnO_(4) to grams. Calculate the percent composition of all the elements in: AgH_(2)PO_(4) Convert 2.4times 10^26 molecules of Al_(2)O_(3) to moles

To convert 400 g of $Mg(CN)_{2}$ into mol:<br /><br />1. Calculate the molar mass of $Mg(CN)_{2}$:<br /> - Molar mass of Mg = 24.305 g/mol<br /> - Molar mass of C = 12.011 g/mol<br /> - Molar mass of N = 14.007 g/mol<br /> - Molar mass of $Mg(CN)_{2}$ = 24.305 + 2(12.011 + 14.007) = 92.32 g/mol<br /><br />2. Convert grams to moles:<br /> - Moles of $Mg(CN)_{2}$ = 400 g / 92.32 g/mol = 4.33 mol<br /><br />The final answer is 4.33 mol.<br /><br />To calculate the number of formula units in 600 g of $NaNO_{2}$:<br /><br />1. Calculate the molar mass of $NaNO_{2}$:<br /> - Molar mass of Na = 22.990 g/mol<br /> - Molar mass of N = 14.007 g/mol<br /> - Molar mass of O = 15.999 g/mol<br /> - Molar mass of $NaNO_{2}$ = 22.990 + 14.007 + 15.999 = 52.996 g/mol<br /><br />2. Convert grams to moles:<br /> - Moles of $NaNO_{2}$ = 600 g / 52.996 g/mol = 11.30 mol<br /><br />3. Calculate the number of formula units:<br /> - Number of formula units = 11.30 mol × (6.022 × 10^23 formula units/mol) = 6.82 × 10^24 formula units<br /><br />The final answer is 6.82 × 10^24 formula units.<br /><br />To convert $8.5\times 10^{27}$ molecules of $KMnO_{4}$ to grams:<br /><br />1. Calculate the molar mass of $KMnO_{4}$:<br /> - Molar mass of K = 39.10 g/mol<br /> - Molar mass of Mn = 54.938 g/mol<br /> - Molar mass of O = 15.999 g/mol<br /> - Molar mass of $KMnO_{4}$ = 39.10 + 54.938 + 4(15.999) = 158.034 g/mol<br /><br />2. Convert molecules to moles:<br /> - Moles of $KMnO_{4}$ = $8.5\times 10^{27}$ molecules / (6.022 × 10^23 molecules/mol) = 141.5 mol<br /><br />3. Convert moles to grams:<br /> - Mass of $KMnO_{4}$ = 141.5 mol × 158.034 g/mol = 22.34 × 10^3 g<br /><br />The final answer is 22.34 × 10^3 g.<br /><br />To calculate the percent composition of all the elements in $AgH_{2}PO_{4}$:<br /><br />1. Calculate the molar mass of $AgH_{2}PO_{4}$:<br /> - Molar mass of Ag = 107.868 g/mol<br /> - Molar mass of H = 1.008 g/mol<br /> - Molar mass of P = 30.974 g/mol<br /> - Molar mass of O = 15.999 g/mol<br /> - Molar mass of $AgH_{2}PO_{4}$ = 107.868 + 2(1.008) + 30.974 + 4(15.999) = 207.88 g/mol<br /><br />2. Calculate the percent composition of each element:<br /> - Percent composition of Ag = (107.868 / 207.88) × 100% = 51.82%<br /> - Percent composition of H = (2 × 1.008 / 207.88) × 100% = 0.97%<br /> - Percent composition of P = (30.974 / 207.88) × 100% = 14.88%<br /> - Percent composition of O = (4 × 15.999 / 207.88) × 100% = 31.00%<br /><br />The final answer is:<br />- Ag: 51.82%<br />- H: 0.97%<br />- P: 14.88%<br />- O: 31.00%<br /><br />To convert $2.4\times 10^{26}$ molecules of $Al_{2}O_{3}$ to moles:<br /><br />1. Calculate the molar mass of $Al_{2}O_{3}$:<br /> - Molar mass of Al = 26.982 g/mol<br /> - Molar mass of O = 15.999 g/mol<br /> - Molar mass of $Al_{2}O_{