One could look back on the experiment in Module 1 with the sugar water The

# One could look back on the experiment in module 1

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experiment in Module 1 with the sugar water. The “steam” from boiling the water took up a great space or volume than the droplets of condensation that collected on the bottom of the second pan that held the ice cubes)b)an increase in pressure. (3 points)An increase in pressure would result in a decrease in volume as the space between the molecules would decrease with the increase in pressure which would lead to a smaller volume being taken up by the same amount of molecules. As the increase in pressure does not actually compress the gas molecules themselves, it simply decreases the amount of space between each molecule thereby leading to a decrease in the volume the same amount of gas would take up when under an increase in pressure. 3.Calculate the molar mass of the following compounds: (16 points)a)NaCl Na = 22.99amu + Cl = 35.45amu = 58.44amub)CaBr2 Ca = 40.08amu + Br2= 159.8amu (79.90 x 2) = 199.88amuc)K2S K2= 60.2amu (30.10 x 2) + S = 32.07amu = 92.27amud)H2SO4 = H2= 2.02amu (1.01 x 2) + S = 32.07amu + O464amu (16amu x 4) = 98.09amue)Mg (NO3)2 = (N= 14.01amu + O3= 48amu [16 x 3] = 62.01amu) x 2 = 124.02amu + Mg = 24.30amu = 148.32amuf)Al(C2H3O2)3= (C2= 24.02amu [12.01 x 2] + H3= 3.03amu [1.01amu x 3] + O2= 32amu [16amu x 2] = 59.05amu) x 3 = 177.15amu + Al = 26.98 = 204.13amug)Sr3(PO3)2 = (P = 30.97amu + O3= 48amu [16amu x 3] = 78.97amu) x 2 = 157.94amu + Sr3= 262.86amu [87.62amu x 3) = 420.80amuh)(NH4)2CO3 = (N = 14.01amu + H4 = 4.04amu [1.01amu x 4] = 18.05amu) x 2 = 36.10amu + C = 12.01amu + O3 = 48amu [16amu x 3] = 96.11amu4.Do the following calculations, applying unit analysis to find the appropriate units for the answers. (12 points)a)25 m/s x 25 s = 625 meters (d = t x v)b)95 km/h x 15 s x 1h/3600 s x 103m/1 km =(95km / 3600s = 0.0264km/s) 0.0264 km/s x 15 s = 0.396km x 10-3= 396 meters (D = t x v)
c)25 km x 106mm/km x 106mm/km = 25000km = 2.5km x 104d)525 m 15 m/s = 35 seconds (T = d/v) e)25 mol x 18.0 g/mol = 450g = 4.5 x102gf)2.5 g/L x 22.4 L/mol = 56g = 5.6 x 101/mol