CHEM 1201 (Basic Chemistry) – Exam 2 Form 1 | Trusted LSU Exam Guide, Exams of Chemistry

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Exam 2 Form 1 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) How many Zn atoms are contained in 922 g of Zn? A) 5.90 x 1025 Zn atoms B) 7.09 x 1021 Zn atoms C) 8.49 x 1024 Zn atoms D) 4.18 x 1024 Zn atoms E) 4.27 x 1022 Zn atoms 922 gZn x 1molZn x ( 022 x 1073 Zn atoms J 1molZn 65.39 mol ) = 8.49 x 10*4 Zn atoms 2) What is the empirical formula of a substance that contains 5.28 g of C, 1.11 g of H, and 3.52 g of O? A) CgH50 B) C3H404 Calculate moles of each element, then divide by the = neyo smallest number of moles: Oxygen (OQ) E) None of these 5.289 Moles of C = ==—4 ~ 0.440 mol “0006 _ nol 0.220 mol 141 1.10 mol H Moles of H = —=~2— = 1.10 mol 74 — 1.008 g/mol 0.220 mol 0 3.52 0.220 mol 0 Moles of 0 = ——£— x 0.220mol —™ = 16.00 g/mol 0.220 mol 0 Empirical formula: C2Hs0 3) What mass (in kg) does 4.77 moles of nickel have? A) 0.280 kg B) 0.352 kg Mass = moles xX molar mass C) 0.632 kg D) 0.122 kg 58.693gNi 1kg Ni E) 0.820 kg m = 4.77 mol Nix * Teagg nim 0-280 ko 4) What is the coefficient for oxygen when the following equation is balanced using the lowest, whole- number coefficients? —— 63Hg92 (I) + 02(9) > CO2 (g) + H20 (I) A)1 B)5 C)3 D)7 E) None of these Balance reactants > Products 7 C3H,02, + 30 > 3C0, + 3H,0 Count oxygen in products: 9 Twa oxygen comes from C3Hg02, which requires 9-2=7 more oxygen 7 502 = 3.50 232 2 Multiply all coefficients by 2 to remove fraction. 2C3;H¢O2 + 702 > 6CO, + 6H2,0 5) How many molecules of HCI are formed when 90.0 g of water reacts according to the following balanced reaction? Assume excess ICI3. 2 IClg +3 H2O = ICI + HIOg + 5 HCI A) 3.00 x 1024 molecules HCI B) 9.00 x 1024 molecules HCI C) 5.00 x 1025 molecules HCI D) 5.00 * 1024 molecules HCI E) 6.00 x 1024 molecules HCI 1molH,0 S5molHCl 6.022 x 1073 molecules 90.0 g H20 « = 24 18.019 H,0" 3molH,0" Lmol HCl 5.00 x 10°*molecules HCL 6) How many moles of BCI3 are needed to produce 5.00 g of HCI(aq) in the following reaction? BCl3 (g) + 3 H20 (I) > 3 HCI (aq) + B(OH)3 (aq) A) 0.0457 mol B) 0.411 mol C) 21.9 mol D) 0.137 mol Moles of BCl; = E) None of these ImolHCl 1mol BC; 5.00 g HCl* sore Gg HCl” 3molHcl ~ ° 9457 mol 10) Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) + N2 (g) — 2Li3N (s) In a particular experiment, 1.00-g samples of each reagent are reacted. The theoretical yield of lithium nitride is g. A) 0.84 B) 1.01 C) 2.50 .(1mol Li, 2molLliz;N | 34.83 9 Li,N 1.00 g Li ( ) x 6 mol Li 1 mol Liz;N Theoretical yield of Li,N D) 1.67 E) 5.0 &34gui)* = 1.67 9 Li;N 1.009 N, ( 1mol N, ) ZmolLizN | 34.83 g Li;N x = 2.49 g LiN 28.02 gNz/” 1molN, ~ 1molLi,N aes Liis the limiting reagent 11) How many grams of NaOH (MW = 40.0) are there in 500.0 mL of a 0.250 M NaOH solution? A) 114 B) 14.0 Mass of NaOH = Volume (L) x Molarity (M) x Molar Mass (45 C) 0.00219 ‘mol D) 0.125 E) 5.00 = 500.0 mL NaOH + 1LNa0H -250molNaOH 40.0 g NaOH 1000 mL NaOH 1LNaQH ~~ 1molNaOH =5.00g 12) What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 0.871 mol CuCl2 in water and diluting to 259 mL? A) 0.00336 and 3.36 B) 3.36 and 3.36 C) 0.00336 and 0.00336 D) 0.297 and 0.113 E) 3.36 and 6.73 Dissociation equation for CuCl,: CuCl,(s) > Cu2* (aq) + 2Cl- (aq) For every 1 mole of CuCl. that dissolves 1 mole of Cu” and 2 moles Cl are produced. moles of solute Molarity (M) = volume of solution (L) 0.871 mol CuCl, 1mol Cu?* 2+] [eu*] 0.259L 1molCuCl, =3.36M Cu* 0.871 mol CuCl, 2mol Cl- []=——Q 3501 * mol CuCl; = 6.73 MCL" 13) What volume (mL) of a 5.45 M lead nitrate solution must be diluted to 820.7 to make a 1.41 M solution of lead nitrate? A) 3170 B) 0.00471 C) 0.00936 D) 212 v= (1.41 M)(820.7 mL) E) 6310 5.45 M Vy =212.4mL = 212 mL 14) A solution is prepared by mixing 15.0 mL of 0.100 M HCI and 5.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution? = ne cr|= (Volume(L)of NaCl x M of NaCl) + (Volume(L)of HCL x M of HCl) Cc) 8.00 [er] = Total Volume (L) D) 1.25 200 mol NaCl 100 mol HCL E) 0.0250 ep 005 L NaCl (Ter) + 0.0151 NaCl (a —) (005 L+.015L) =0.125M 15) How many grams of CaCl2 are formed when 15.00 mL of 0.00237 M Ca(OH)2 reacts with excess Cl2 gas? 2 Ca(OH)2(aq) + 2 Cla(g) > Ca(OCl)2(aq) + CaCla(s) + 2 H20(1) A) 0.0507 g B) 0.00 197 g C) 0.00789 g D) 0.00 394 g E) None of these Mass of CaCl, 1LCa(OH), __ 0.00237 moles Ca(OH), _1molCaCl, 110.98 g CaCl, _ 15.00 ml Ca(OH) * 7500 mL Ca(OH)a” TL Ca(OH), * ZmolCa(OH),” imolCact, ~ 00197 9 Cache 16) How many milliliters of 0.260 M Na2S are needed to react with 40.00 mL of 0.315 M AgNO3? Na2S (aq) + 2 AgNO3 (aq) — 2 NaNO3 (aq) + Ag2S (s) A) 66.0 mL B) 48.5 mL C) 96.9 mL D) 24.2 mL E) None of these 1L AgNO; -315molAgNO, 1mol Na,5 1LNa,S 1000 mL Na,S * * * * 1000 mL AgNO; 1LAgNO; 2mol AgNO; .260 mol NaS TL Na2S 40.00 mL AgNO; « =24.2mL Na,S