Question

Problem 7.37 - Clinical Applications Calculate the molar mass for each of the following Part A Al(SO4), antiperspirant Express your answer to two decimal places and include the appropriate units. ANSWER Part B NaHCO, baking soda Express your answer to two decimal places and include the appropriate units ANSWER Part C C, H, NO, paracetamol used to treat pain and fever. Express your answer to two decimal places and include the appropriate units. ANSWER Problem 7.47 Calculate the number of moles in 250 g of each of the following Part A Express the number of mole to three significant figures. ANSWER: mole Part B N2

Answer 1

Answeri calculate molar mass. Part A Al2(Soub mass of At: 26.982 mass of S: 62066 & mass of 0:15-999 2X AI + 3x S + 12x0 2x 26.982 + 3x 32-066 + 12 X 15-999 = 53.964 +96.198 + 191988 | Al2(SO4)₂ = 342 15 1 g/mol Part B NaHco3 mass of Na : 22.989 mass of H:1.0079 mass of c: 12-0107 mass of o: 15.999 NatHtc +(3+0) 22.989 + 1-0079 + 12-0107 + (3 x 15-999) = 22.989 +1.0079 + 12-0107 + 47.997 = 36.0076 + 47.997 NaHCO3 = 84.005 g/mol

pool . (& Hg NO2 mass of c= 12.0107 mass of H:1-0079 mass of N. 14.0067 mass of 0 = 15-999 (8x0) +(9XH) + N + (2x0) = (8x 12.0107) + (9 x 1.0079) +14.0067 + (2x15.999) = 96.0856 + 9.0711 + 14.0067 31.998 | (8 Hg No 2 = 151-16 I g/mol problem 7.47 Calculate the number of moles in 25g of each of the Following: part A: molar mass of B = 10.8 11 g/mol. mass = number of moles x molar mass. " number of moles = mass 25 molar mass = 2.312 moles 10.811 Number of moles OF B = 2.312 moles

Past B N2 molar mass of N = 14.0067 N2 = 2x 14.0067 = 28.0134 Number of moles of Na= mass molar mass of N2 = 25 28.0134 Number of moles of N₂ = 0.892 moles part : In (OH)3 molar mass of In = 114.818 0= 15.ggg H=1-00794 Int (3x0) +(3×H) = 114.818 + (3x 15-999) + 3 x 1.00794) = 114.818 + 47.997 + 3.0238 = 165.83 g/mol. Number of moles of Incora3 = - 25 165.83 Number of moles of IncoH)₂ = 0.150 moles. Part o - و کم A۱ A1 = 26.982 s= 32.066 (2x 26.982) + (3x 32.066) Ali z = = $3.964 + 96.198 = 150.162 Alzl3

Number of moles of Alas3 = 25 150 162 Number of moles of Al263 0.166 moles Part : (3 48. propane = 12.0107 H= 1.0079 (3 Hg = (3x12.0107) + (8 * 1.0079) = 36.0321+ 8.0632 = 44.0953 glossol Number of moles of (348 = 25 44.0953 = 0.566 moles. Number of moles of (348 = 0.566 moler Problem: 7.59 Answer: mole of 502= 0.540 mol moles of carbon needed & soz moles of a needed to react with certain moles of is calculated using mole ratio: 5 mole 2 molsoL moles of c = mol'c 0.540 moles of so, X 5 . 2 mol soz

moles of c= 1.35 molcl to react " Number of moles of c are needed with 0.540 mole so2 = 1.35 mole