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Sample Test 1

# Sample Test 1 - CHEM 101 SAMPLE EXAM items in BOLD...

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CHEM 101 SAMPLE EXAM: items in BOLD characters are the ANSWERS. If I were you, i’ll answer first before looking at the answers. 32 1) The species with the same number of electrons as S is 16 @ 35 - Cl 17 @ 31 P 15 @ 34 + S 16 @ 35 2- S 16 @ 40 2+ Ar 18 @ 5 32 40 40 2+ S has 16 electrons. Ar has 18 electrons so Ar has 18-2 = 16. 16 18 18 2) A student wishes to obtain 125g of NaCl(s) by evaporating to dryness a quantity of seawater which is 3.5% NaCl by mass. Seawater has a density of 1.03 g/cm^3. How many liters of seawater are required? @ 12.3 2.0 3.7 3.5 34 4 3.5% of mass of seawater required = 125g so M(seawater) = 125/.035 = 3571g 3571g / 1.03g/cm^3 = 3467cm^3, 3467cm^3 / 1000cm^3/L = 3.5L 3) Four of the following might be either empirical or molecular formulas, but one can only be a molecular formula. That one is @ NH2 N2O N2O4 P2O3

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C12H22O11 The empirical formula for N2O4 is NO2. 4) The atom/ion which contains 12 protons, 14 neutrons, and 10 electrons is @ 14 Ne 10 26 2+ Mg 12 14 Mg 12 26 2+ Si 14 26 Mg 12 A Here Z = 12 = atomic no. or # of protons The symbol is Symb . A = Z + N(# of neutrons) = 12 + 14 = 26 Z charge (nñ) = Z - # of e-'s = 12 - 10 = 2 Symb. for Z = 12 is Mg. 5) The number of hydrogen atoms in 1.0mg of C3H8 is @ 1.1E20 8.2E22 8 1.8E-4 6.5E23 1E-3g / 44.03(MW of C3H8) = 2.27E-5 mol(C3H8), 8 * 2.27E-5 = 1.82E-4 mol(H) 6.022E23(Avogadro's no.) * 1.82E-4 = 1.1E20 6) Naturally occurring copper (Cu) consists of two isotopes: one of these has a mass of 62.91 amu and a natural abundance of 69.09%. What is the atomic mass (amu) of the other isotope? @ 61.91 64.98 29.0 69.09 63.91 63.55(MW of Cu) = .6909 * 62.91amu + (1 - .6909) * X amu X amu = 20.085 / .3091 = 64.98 7) What is the mass percent of magnesium in magnesium nitrate? @ 43.96 28.17 16.39
63.45 20.91 mass % = 100 * 24.31(MW of Mg) / 148.33[MW of Mg(NO3)2] = 16.39 8) Compound X3O4 is 72.03% X by mass. The atomic mass of X is @ 30.98 50.95 9.013 19.00 54.94 16.0(MW of O) * 4 = (1 - .7203) * MW of X3O4 so 64.0 /.2797 = 228.82 228.82(MW of X3O4) = 3 * MW(X) + 4 * 16.0(MW of O) so 3 * MW(X) = 164.82 MW(X) = 164.82 / 3 = 54.94 9) The reaction a C2H8N2 + b N2O4 --> c CO2 + d N2 + e H2O is the basis for the space shuttle thruster rockets. The sum of the smallest integer coefficients (a+b+c+d+e) in the balanced equation is @ 24 12 8 10 11 C2H8N2 + 2 N2O4 --> 2 CO2 + 3 N2 + 4 H2O thus (1 + 2 + 2 + 3 + 4) = 12 10) In the table below, select the correct entry for the compounds which are formed when the element reacts with O2(g).

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Sample Test 1 - CHEM 101 SAMPLE EXAM items in BOLD...

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