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Course: CHEM 125, Winter 2012
School: Ohio State
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Word Count: 947

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Zellmer Time: Dr. 30 mins Chemistry 125 Winter Quarter 2012 Pre-quiz Name KEY Thursday January 5, 2012 Rec. TA/time. 1. (7 pts) The amount of NO2 in the air on a particular day is 1.55 x 10!6 kg/m3. How many pounds of NO2 are present in 2.00 ft3? (1 in = 2.54 cm, 1.000 lb = 453.6 g) (You MUST use dimensional analysis to receive full credit!) 2. (8 pts) A quantity of heat equal to 600.0 cal is ADDED to...

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Zellmer Time: Dr. 30 mins Chemistry 125 Winter Quarter 2012 Pre-quiz Name KEY Thursday January 5, 2012 Rec. TA/time. 1. (7 pts) The amount of NO2 in the air on a particular day is 1.55 x 10!6 kg/m3. How many pounds of NO2 are present in 2.00 ft3? (1 in = 2.54 cm, 1.000 lb = 453.6 g) (You MUST use dimensional analysis to receive full credit!) 2. (8 pts) A quantity of heat equal to 600.0 cal is ADDED to 40.0 g of water at an initial temperature of 35.0oC. Calculate the FINAL temperature (in oC). (The specific heat of water is 1.00 cal/gCC) Show your work. q=m heat C c C )T (mass) C (sp. heat) C (temp. change) q = + 600.0 cal (heat is added or required so q is positive; endothermic) Want final temp. so must find )T first q 600.0 cal )T = -------- = ----------------------------- = + 15.0C (40.0 g)(1.00 cal/gCC) (3 s.f.) mCc )T = Tf & Ti Tf = Ti + )T = 35.0C + (15.0C) = 50.0C (3 s.f.) Copyright R. J. Zellmer, January 5, 2012 35.0 (only known to tenths) + 15.0 -----50.0 (to tenths place) 3. (3 pts) Correctly name or write the formulas for the following compounds. Also, are they ionic or molecular? Chromium (VI) oxide Cr6+ O2& ==> The roman numeral gives the charge on the chromium ion (+6). CrO3 IONIC For future reference, you should know how many particles (ions) you get when you put an ionic substance in water. First of all it dissociates (comes apart). Thats the easy part. You also need to remember that these polyatomic ions stay together in solutions (i.e. act as a single particle). H2 O CrO3 (s) ----------> Cr6+ (aq) + K3PO4 3 O2& (aq) potassium phosphate IONIC H2 O K3PO4 (s) ----------> 3 K+ (aq) + P2O5 4. (you get 4 particles in solution per formula unit) PO43& (aq) diphosphorus pentoxide (you get 4 particles in solution per formula unit) MOLECULAR (2 pts) Determine the oxidation number of the underlined element in the following compound. (Show all work.) Chg. on HSO4& is -1 HSO4& 5. 1(+1)H + XS + 4(-2)O = -1; XS ! 7 = -1; XS = +6 (2 pts) In the following reaction, which element is being oxidized and which is being reduced? N: reduced HNO3: oxidizing agent Copyright R. J. Zellmer, January 5, 2012 Cu: oxidized Cu: reducing agent 6. (2 pts) Internal energy, E, is a state function. What does this mean? Internal energy is a state function. A state function depends only on the present conditions and not how the system got to its present state. It doesnt depend on the path taken. Other examples of state functions are potential energy, temperature, enthalpy, pressure, volume and others. A difference in state functions is also a state function (i.e. )E = Ef - Ei and since internal energy is a state function the change in internal energy for a process is state function). 7. (3 pts) The standard enthalpies of formation, )Hf, at 25C for SO2(g) and SO3(g) are -296.9 kJ/mol and -395.2 kJ/mol, respectively. What is the enthalpy of reaction, )H is AND the reaction endothermic or exothermic? 2 SO2(g) + O2(g) )Hf (kJ/mol) H - 296.9 0 = nH v 0.0 0 products 2 SO3(g) ()Hf for elements in their standard states is zero by definition) - 395.2 products mH 0 reactants reactants )H = [(2 mol)(- 395.2 kJ/mol)] - [(2 mol)(- 296.9 kJ/mol) + (1 mol)(0.0 kJ/mol)] = [-790.4 kJ] - [-593.8 kJ] = - 196.6 kJ Exothermic (a negative )H; energy is released, as a product) 8. See next page. 9. (1 pt) What volume (in mL) of a 5.0 M NaCl solution is needed to make 500.0 mL of a 2.0 M NaCl? Dilution problem so use M1 = 5.0 M M2 = 2.0 M M2*V2 = M1*V1 V1 = ? mL V2 = 500.0 mL M2*V2 (2.0 M) * (500.0 mL) V1 = ----------- = --------------------------- = 200.0 mL = 2.0 x 102 mL (5.0 M) M1 Copyright R. J. Zellmer, January 5, 2012 8. (12 pts) Consider the following molecules and list their molecular shapes (NOT the electron domain geometries) and whether they are polar or nonpolar. Provide a short explanation for your choices. 1) H2S bent (. 109.5 angles) 4 things around S and not all identical (2 H and 2 lpe&) .. .. .. Polar (like H2O:, H2Se:, :SCl2, etc.) 2) BH3 trigonal planar (= 120 angles, exactly) 3 things around B and all identical (3 H and NO lpe!) Nonpolar (as is BCl3, BF3 AlCl3 etc.) 3) NF 3 trigonal pyramidal (. 109.5 angles) 4 things around N and not all identical (3 F and 1 lpe!) Polar (like :NH3,:PH3, :PCl3, etc.) 4) CH3Cl tetrahedral (. 109.5 angles) All 4 atoms on C are not identical Polar (as is CHCl3, CH3F, CH2F2, ect.) NOTE: The 5 basic symmetric molecular geometries (linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral) can give nonpolar but only if all surrounding atoms (or groups of atoms) are identical to each other. See Tables 9.2 and 9.3 on pages 347 and 350 in the textbook. * NOTE: If there are 1 or more lpe- on the central atom the molecule will generally be polar. There are two shapes that can be exceptions to this rule. (See Table 9.3 on page 350 of the textbook.) The linear shape resulting from the trigonal bipyramidal e--pair geometry will be nonpolar if the 2 atoms around the central atom are identical to each other (e.g. XeF2 has 3 lpe- in a trigonal planar geometry around the Xe and the 2 F atoms linear to each other). The square planar shape resulting from the octahedral e--pair geometry will be nonpolar if the 4 atoms around the central atom are identical to each other (e.g. XeF4 has 4 F atoms in a square planar arrangement with 2 lpe- linear to each other, 1 above and 1 below the square planar arrangement). For BH3 (#2) all atoms surrounding B are identical so the molecule is nonpolar. H2S and NF3 (1 & 3) with lpe! on the central atom are polar. In general, molecules that have 1 or more lpe! on the central atom are polar. (See discussion above about exceptions.) For CH3Cl (#4) all surrounding atoms are NOT identical so the molecule is polar. Copyright R. J. Zellmer, January 5, 2012
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