Question

COURSE NUMBER: CHEM 1111 Chemistry: Chemical Quantities and Reactions Spring 2020 Homework Set 5 1. Calculate each of the following: a. number of Li atoms in 4.5 moles of Li b. number of CO, molecules in 0.0180 mole of CO, c. moles of Cu in 7.8 x 1021 atoms of Cu d. moles of C,H, in 3.75 X 102 molecules of CH6 2. Calculate each of the following quantities in 0.185 mole of C6H 40: a. moles of C b. moles of o c. atoms of H d. atoms of C

1 and 2 please.

Answer 1

= mole - Crivenmass = Number of entities Molar mass Anagadro Number (6.022 x 1023) particles. = 1 mole of any entity contains 6.022 X 1023 Now, as moles of Li = 4.5 moles in number of Li atoms = 4.5 moles x6•022 X 1023 = 27.099 X1023 = 2.7099 X 1024 atoms. b) mole of CO2 = 0.0180 mole .. number of co molecules = 0.0180 x 6.022x10² = 0.108396 X 1023 = 1.08396 X 1022 molecules c) Number of atours of Cu = 7.8 x 1021 atomo . moles of Cu = Number of atoms = 7.8 X 1021 Avogadro No. 6.022 x 1023 = 1.295 x 10-2 = 0.01295 mole d) Number of C₂H6 molecules = 3.75 x 1023 molecules : moles of C₂ H6 = Number of molecules - 3.75 x 1023 Avogadro No. 6.022 X 1023 = 0.622 mole ② Mole of C6H140 = 0.185 mole So, 1 mole of Co H140 molecule contains carbon atom , 14 moles of hydrogen atom of Oscygen atom. 6 moles of and 1 mole

a) moles of c = 6x 0.185 = loll moles b) moles of o= 1 x 0.185 = 0.185 mole c) moles of H = 14 x 0.185 = 2.59 moles in atoms of H= moles & Amagadro Number = 2.59 x 6.022x1023 = 15.596x 1023 = 1.5596x 1024 atours of H. dy moles of C - boll moles atoms of C = moles & Auogadro Number - loll x 600 22X1023 = 6.68442 x 1023 atoms of c
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