Question

m105%20Makeup%20Exam%20(2).pdf 31) 31) To convert a given number of moles into the number of atoms, one would multiply by which of the following factors? A) 1.66 x 10-24 atoms/I mol B) 1 mol/6.022 1023 atoms C) molar mass D) 6.022 * 1023 atoms/1 mol E) 1 mol/1.66 x 10-24 atoms 3 2) 32) How many atoms of sulfur are present in a 155 g sample of sulfur? [Molar mass: S 32.06 g/mol) A) 6.02 x 1023 atoms B) 2.91 x 1023 atoms C) 2.01 x 1023 atoms D) 2.91 x 1024 atoms E) 3.01 x 1023 atoms 33) 33) What is the formula mass of carbon dioxide? [Atomic masses: C, 12.01 amu; O, 16,00 amu] A) 44,01 g B) 44,01 amu C) 28.01 g D) 28.01 amu E) 44.01 mol 34) 34) What is the mass of a 0.0200 mol sample of nicotine? [Formula mass: nicotine, 162.2 amul A) 8.11 x 103 g B) 32.48 C) 3.24 s D) 1.23 x 10-4 g E) 0.308 g 35) 35) The chemical equation below is not balanced. What is the coefficient of iron when the equation is balanced with smallest whole number coefficients? Fe(s) + O2(8) - Fe2O3(3) A) 4 B) 8 01 D) 3 E) 2

Answer 1

☺ * 1 3 ☺ G 022 81023 atomg I 1 mol No. of atom = 6022 X 1023 atuny | 1 mol. moler at S= mass = ISSg , = 4.83 mol molarmas - 32.06gmort Number of Satem= 6.022x10 23 atery x 4.83 mal. Imos = 2.91 x 1024 atong (0) Coz Pormela man = atomic nan of ct axatenne mass of o 12.0lamu + 2x 16oo amu = 44.01 amu ( B) man of mictoine = male x formula mass = 0.0200 male & 162.2 amy - 3.209 (c) 4Fe +30219) 2Fe2O3 А) wч - А) 6 (15)