Exersises solutions for chemistry 1st year course, Exercises of Chemistry

Download Exersises solutions for chemistry 1st year course and more Exercises Chemistry in PDF only on Docsity! CHM1311F-DGD_Oct20 TA: Anastasiia Suprun Prof. Dr. Sandro Gambarotta Question 1 The chemical formula for aluminum sulphate is Al2(SO4)3. (a) Compute its molar mass. (b) Compute the number of moles contained in 25.0 g of this compound. (c) Determine its percent composition. (d) Determine the mass of this compound that contains 1.00 mole of O. Se + Cr(OH); — Cr+SeO;?_ (basic conditions) Se — SeO;> (1) Se is balanced; (2) add 3 H2O on the left; (3) add 6 H2O on the right and 6 OH on the left; (4) add 4 e& on the right: Se + 3 HO + 6 OH” —> SeQ;? + 6 HO+4e Cancel duplicated species: Se + 6 OH” — SeO;7 + 3 H20+4e Cr(OH)3 — Cr (1) Cr is balanced: (2 and 3) by inspection, add 3 OH™ on the right; (4), add 3 e& on the left: Cr(OH)3 + 3 e& — Cr +3 OH Multiply the first reaction by 3 and the second reaction by 4, and then add: 3 [Se +6 OH” > SeO,* +3 H,O0+4e°] + 4 [Cr(OH), +3 e > Cr+3 OH] 3 Se+4 Cr(OH), +6 OH > 3 SeO,” +4 Cr+9H,O Question 3 A student prepares a solution by dissolving 4.75 g of solid KOH in enough water to make 275 mL of solution. (a) Calculate the molarities of the major ionic species present. (b) Calculate the molarities of the major ionic species present if 25.00 mL of this solution is added to a 100 mL volumetric flask, and water is added to the mark. The solution process is KOH (s) — K~ (aq) + OH™ (aq). Each mole of solid generates 1 mole of each ion: (a) Molarity is found using the equations, c =F and n= i: M = 39.098 g/mol + 15.999 g/mol + 1.0079 g/mol = 56.11 g/mol 3 en 2 -( 758.) Imol \{10? mL eugisppawies opt. V \275 mL )\ 56.112 LE: a (b) In a dilution, the number of moles of solute remains constant whereas volume Increases, so c;/’;= cil’). The starting volume is 25.00 mL and the final volume is 100.00 mL: _ eV, _ (0.308 M)(25.00 mL) Ce = 0.0770 M= Cqr =E V, 100. mL OH Question 5 Name the hybrid orbitals formed by combining the following sets of atomic orbitals: (a) one 2s and two 2p orbitals; (b) one 3s, three 3p, and one 3d orbital; and (c) one 4s and three 4p orbitals. The name of a hybrid orbital set includes the letter of the atomic orbitals, and the number of each type, used to make the set: (a) sp’; (b) sp’at and (c) sp’. Question 6 Draw the Lewis structures for the two possible arrangements of the N20 molecule, NNO and NON. Experiments show that the molecule is linear and has a dipole moment. What is the arrangement of atoms? Justify your choice. Use the Lewis structure of the molecule: (a) sp” hybrids; (b) sp? hybrids; (c) sp° hybrids (d) sp° hybrids. Question 8 Both PF; and PFs are known compounds. NF; also exists, but NF; does not. Why is there no molecule with the formula NFs? The Lewis structures of molecules with formula XF3 show octets around the inner atom and FCy = 0, making them stable. Compounds with formula XFs also have FCy = 0 but have five electron pairs associated with the inner atom. This is possible for phosphorus, a third-row element that has d orbitals available for bonding. It is not possible for nitrogen, a second-row element that lacks valence d orbitals: 7 . . a 2c =n! >—— x — TN Py = il = a | u .: Question 10 Identify the hybridization of the underlined atom in the following species: (a) SF’; (b) (CH3).CO; (0) BF, ’; and (d) HCCCHs. The steric number of an inner atom uniquely determines its hybridization. Use the Lewis structure of the molecule to find steric numbers: () <fFs ®) # © 2%. (a) | > - ww” ae of H oe o* La a al “* - — Ni FJ ‘er ae | i “sro ae Be (a) SN = 6, sp’ d hybrids; (b) SN = 3, sp’ hybrids; (c) SN = 4, sp’ hybrids; and (d) SN = 2, sp hybrids. Question 11 Use molecular orbital diagrams to rank the bond energies of the following diatomic species, from weakest to strongest: H», H2 , and Ho2. The orbital energy-level diagram for first-row diatomic species has just two levels, each of which can hold two electrons. The relative bond strengths can be deduced from the bond orders, which are calculated using Equation 7-1 of your textbook: to tee —e “fe the +e * 2+ He, He, He, BO=0 BO=1/2 BO=1 From weakest to strongest, the order is He2 < Hes" < Heo”. Question 13 In each of the following pairs, which has the stronger bond? Use orbital configurations to justify your selections: (a) O2 or O7+; (b) CO or CO; and (c) Fo or Fo". The stability of a diatomic molecule is determined by the number of bonding and antibonding electrons. To compare species, determine how many valence electrons each species possesses, and then place the electrons in the available orbitals, following the Pauli and aufbau principles. Here are the configurations for the three neutral species: fy) O, (12 e's ib) CO 16 &a fo Fy 14 es) o* ot — — r* — = + n* —..—_ ge qh ah ne + + tot Ho He + = we qb re = oe + + + (a) To make O2", an antibonding electron is removed from Oz, so O2" has a stronger bond. (b) An electron added to CO to produce CO’ occupies an antibonding orbital, so CO has a stronger bond. (c) An antibonding electron is removed from F2 to produce F2", so F2" has a stronger bond.