Question

Question 3 7 pts Calculate the root mean squared velocity of Oxygen gas, O3 at 96°C, in meters per second. Question 4 7 pts Dissolving 5.44 grams of an impure sample of calcium carbonate in hydrochloric acid produced 0.334 liters of carbon dioxide measured at 20.0 °C and 792 mmHg. Calculate the percent by mass of calcium carbonate in the sample. CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) +H20(1) + CO2(g)

Answer 1

Angwer Coiven Сосоз mass 5.449 = 0.334 L Volume con produced Temperature 20°C (20+273)K = 293K Pressure 792 mmH mmHg 792 mmig latm = 1.042 atm 760mmHg % ma99 Of Сасоз ? chemical eq CaCO36) 2 HCI@2) - Caci, (qq)+ H20() +20 (3) solution ideal calculate moles cue using gas ܕܘܘܠ p Pressure Tova NRT IRT Volume no cf moles pv = gas constant Co. 0821L atm Imolk) RT T - Temperature of noof moles con 1.042 atm xo.334. 0.0145 mol x 2935 O 0821 Latm/meik COR From the balanced chemical equation, I mol 1 mol Caco 3 of CaCo3 moles 0.0145 mol CaCO3 of molas mags of CaCO3 100.09 glmoi X nolaama9S mol of Cacoy mas9 of CaCO3

= CaCO3 0.0145 mol X 100.09.9lmol mags of 1.459 Sample 이 이 mass of CaCO3 1.459 CaCo3 x100 5449 caco, IT 26.65% criven Temperature 96°C 11 Rms velocity of Solution R . 8.314 3 /molk 3 RT Nams M T Temperature . molas mas 9 in kg/mol 3 qĞc (96+273)* = 369K molas mass of Oz 48 g/mol { Iglmol 0.001kg/mol = 0.048 kg/mol Vams = Nom 3*8.314 YhXmék.* 369 K 1.048 139mor 11 kg m.m » kg 191741.62 m/s2 De 437.88m/s