Test 2 Solutions

# Test 2 Solutions - 1) C(graphite), H2(g), and O2(g) are the...

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Unformatted text preview: 1) C(graphite), H2(g), and O2(g) are the standard states for those elements and must be used when calculating ∆ Hf. 2) ∆ E = q + w. System does work = energy out, thus w = -125 J. Heat added to system = energy in, thus q = +75 J. ∆ E = +75 + (-125) = -50. J. 3) q = n(mol) x C(J/¯C-mol) x • T(¯C) n = 162g / 63.55g/mol = 2.549 mol 2.549 x 25 x 74.1 = 4700 J = 4.7 kJ 4) 3.00g / 12.01g/mol = 0.2498 mol of C ∆ Hf (kJ/mol) = -98.3 kJ / 0.2498mol = -394 kJ/mol 5) 4s2, 3d6 is correct. The 6 d electrons are arranged in the 5 levels like: ud u u u u-- -- -- -- -- where u = spin up and d = spin down, thus there are 4 unpaired electrons. 6) 4s^2, 4p^6, 4d^10, 4f^14. 2 + 6 + 10 + 14 = 32. 7) Cs(s) + H2O(l) ---> CsOH(aq) + 1/2 H2(g) The heat of this reaction often ignites the H2 gas causing an explosion! 8) E = h(Planck's constant) x v(frequency) thus energy is proportional to frequency 9) In the 3rd row Mg < P < Si, and in the same column Mg < Be so Mg is the smallest overall. 10) In an isoelectronic series, the more negative (or less positive) ions are always larger. 11) 2 H2(g) + O2(g) ---> 2 H2O(g) 2 x (-242) = -484 C(graphite) + O2(g) ---> CO2(g) 1 x (-394) = -394 CO2(g) + 2 H2O(g) ---> CH4(g) + 2 O2(g) -1 x (-803) = +803-----------------------------------------------------------------...
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## This homework help was uploaded on 04/17/2008 for the course CHE 101 taught by Professor Churchhill during the Spring '08 term at SUNY Buffalo.

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Test 2 Solutions - 1) C(graphite), H2(g), and O2(g) are the...

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