Question

MEDGAR EVERS COLLEGE OF THE CITY UNIVERSITY OF NEW YORK Department of Chemistry and Environmental Science CHM 201-002 Exam 2 B November 7, 2019 Dr. Harini Patel Names actuacks Section A: (Circle or bubble the correct answer with PEN) 1 Which solution of a base contains the smallest number of moles of a base? a) 5.0 L of 4.0 M b ) 0.50 L of 4.0 M €) 4.0 L of 0.40 M d) 0.50 L of 0.40 M What is the volume of 0.1220 M HNO, solution required to completely neutralize 17.50 ml of 0.1211M NaOH ? Reaction is: HNO, (aq) + NaOH(aq) - NaNO, (aq) + H;O (aq) a) 17.84 m b) 16.31 ml. c) 15.39 mL d) 17.37 ml Using the following portion of the activity series of elements Mg(8) ► Mg +(aq) + 2e" (strongly reducing) Al(s) → Al(aq) + 3e Cu(s) → Cu+(aq) + 2e- (weakly reducing) Determine which one of the reactions will occur. a) Cu(s) + Ca(aq) b) Al(s). Ca 2+ (aq) b) Al(s) + Cu?" (aq) d) Cu(s) + AP + (aq) In the following combustion reaction of methane gas (CH) with oxygen is given. How many (g) of H20 is produced when 0.533 moles of CH4 reacts with O2? Reaction is: CH. + 20 - CO2 + 2H:0 a) 19.21 g b) 17.21 g c) 9.60 d) 15.23 g 5 How many grams (of NaCl are needed to prepare 250.0 ml of 0.365 M of NaCl. Solution? a) 3.65 g b) 10 25 g c) 5.34 g d) 7.21 g IF 0.500 L of a 3.50 M HCl acid is diluted to 1.00 L, the molarity of the new solution is: a) 1.50 M b)3.50 M )1.40 M d) 1.75 M

Answer 1

(1) Molarity Volume Molarity X Volume (6) 6. 50L of 4.0M Number of moles= (a) 5.0L of 4.0M moles = 4.005.0 moles = 0.50X4.0 = 2.0 Moles = 20 moles (d) 0.50L of 0.40M Moles = 0.5X0.40 = 0. 20 (6) 4.0L 0804.om mous = 480.4 = 1.6 moles Therefore correct option is (a) (2) (MV) HNO3 = (MU) NOCH (0-1220x VINO3) = 0.1241 x17.5 O TomL VHNO3 = 2:11925 0:1220 VHNO3 = 17.370mL Thorlove correct option üld) (3) (a) Cu(s) + cats cag) - > No veaction (6) Al(s) + Catz (aq) -> No te action (c) Al(s) + Cut 2 Lag) - Reaction will occur d) cucs) + Aet3 Cay) - No Recution

(4) CH4 +202 —>602 +2H2O I mole of eHly gives a moles of H₂O 0.533 Moles gives 2x0.533 =1066 mole moles= weight taken. moles mass Weight taken = Moles x molar mass = 1066 x 18 = 19. 188 gms of H₂O will be produced