Answers to homework problems: Chapter 9 part 2. 8) S = F/A = 150 N/ (0.2 m * 0.3 m) = 2500 N/m^2 11) Y = (F/A)/(deltaL/L) = = {600 N/ [pi* (0.001m)^2]}/[(1.3026-1.3000)/1.3000] = 9.42 10^10 N/m^2 32) p = F/A = (75 kg * 9.8 m/s^2)/ 0.0125 m^2 = 5.88 10^4 Pa 33) p = F/A Here F is the component of skiers weight that is perpendicular to the slope F = 90 kg * 9.8 m/s^2 * cos(15 deg) p = 90 * 9.8 * cos(15 deg)/ 0.4 m^2 = 2129.9 Pa Chapter 10 part 1 8 T(F) = 9/5 T(C) + 32 = 9/5 * 39.4 +32 = 102.9 (deg F) 13) Delta(T) = 7 - (-49) = 56 (deg C) Transformation to Farenheit Delta(T) = 56 (deg C) * 9/5 = 100.8 (deg F) 27) a) weight of 1 mole of water (H2O) = 2*1+16=18 g number of moles = total weight/weight of 1 mole = 40g/18g = 2.2 b) weight of 1 mole of H2SO4 = 2*1+32+4*16=98 g number of moles = total weight/weight of 1 mole = 245g/98g = 2.5 c) weight of 1 mole of NO2 = 14+2*16=46g number of moles = total weight/weight of 1 mole = 138g/46g = 3.0 d) use ideal gas law or Example 10.4) 1 mole of any gas occupies 22.4 L of volume --> number of moles = total volume/volume of 1 mole = 56L/22.4L = 2.5 34) 1 mole has a volume of 22.4 L and contains an Avogadro number i.e., 6.02 10^23 molecules. ----> a) 1 L= 1/22.4 of a mole and contains 1/22.4 * 6.02 10^23 = 2.69 10^22 molecules b) 1 cm^3 is 1/1000 of a liter and contains 2.69 10^19 molecules