chem06: gases

<li>A mixture of KCl and KClO3 weighing 1.80 grams was heated; the dry O2 generated occupied 1.40 x 10^2 mL at STP. What percent of the original mixture was KClO3, which decomposes as follows:</li>
</ol>

<p>2KClO3 (s) ----> 2KCl (s) + 3O2 (g)</p>

<p>(A) 28.4%
(B) 37.2%
(C) 42.6%
(D) 63.8%
(E) 72.6%</p>

<hr>

<li>A sample of 35.1 g of methane gas has a volume of 5.20 L at a pressure of 2.70 atm. Calculate the temperature.</li>
</ol>

<p>(A) 4.87 K
(B) 78.1 K (ANSWER)
(C) 46.3 K
(D) 275 K
(E) 129 K</p>

<li>Determine the partial pressure of the hydrogen gas in this mixture.</li>
</ol>

<p>(A) 562 torr
(B) 580 torr
(C) 598 torr
(D) 616 torr
(E) 634 torr</p>

<hr>

<li>Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0deg C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.810 atm.</li>
</ol>

<p>Determine the number of moles of hydrogen gas present in the sample.</p>

<p>(A) 42 mol
(B) 0.82 mol
(C) 1.3 mol
(D) 0.056 mol
(E) 22 mol</p>

<hr>

<li>Oxygen gas, generated by the reaction 2KClO3 (s) ----> 2KCl (s) + 3O2 (g), is collected over water at 27deg C in a 2.00-L vessel at a total pressure of 760. torr. (The vapor pressure of H2O at 27deg C is 26.0 torr.) How many moles of KClO3 were consumed in the reaction?</li>
</ol>

<p>(A) 0.0790 moles
(B) 0.119 moles
(C) 0.0527 moles (ANSWER: Need Explanation.)
(D) 0.0813 moles
(D) none of these</p>

<hr>

<li>A balloon contains an anesthetic mixture of cyclopropane (cp) and oxygen (O2) at 170. torr and 570. torr, respectively. What is the ratio of the number of moles of cyclopropane to moles of oxygen?</li>
</ol>

<p>n<em>cp/n</em>O2 = ?</p>

<p>(A) 0.19
(B) 0.23
(C) 0.30
(D) 0.39
(E) 0.46</p>

<hr>

<li>A gaseous mixture containing 1.5 mol Ar and 3.5 mol CO2 has a total pressure of 7.0 atm. What is the partial pressure of CO2?</li>
</ol>

<p>(A) 1.8 atm
(B) 2.1 atm
(C) 3.5 atm
(D) 4.9 atm
(E) 2.5 atm</p>

<hr>

<li>Consider the following gas samples:</li>
</ol>

<p>Sample A Sample B
S2 (g) O2 (g)
n = 1 mol n = 2 mol
T = 800 K T = 400 K
P = 0.20 atm P = 0.40 atm</p>

<p>Which of the following statements is false?</p>

<p>(A) The volume of sample A is twice the volume of sample B.
(B) The avg KE of the molecules in sample A is twice the avg KE of the molecules in sample B.
(C) The fraction of molecules in sample A having a KE greater than some high fixed value is larger than the fraction of molecules in sample B having KE greater than that same high fixed value.
(D) The mean square velocity of molecules in asmple A is twice as large as the mean square velocity of molecules in sample B.
(E) Assuming identical IMF’s in the two samples, sample A should be more nearly ideal than sample B.</p>

<hr>

<li>At 200 K, the molecules or atoms of an unknown gas, X, have an average velocity equal to that of Ar atoms at 400 K. What is X? (Assuming ideal behavior.)</li>
</ol>

<p>(A) He
(B) CO
(C) HF
(D) HBr
(E) F2</p>

<hr>

<hr>

<li>Which of the following is NOT a postulate of the KMT?</li>
</ol>

<p>(A) Gas particles have most of their mass concentrated in the nucleus of the atom. (ANSWER)
(B) The moving particles undergo perfectly elastic collisions with the walls of the container.
(C) The forces of attraction and repulsion between the particles are insignificant.
(D) The avg KE of the particles is directly proportional to the absolute temperature.
(E) All of the above are postulates of the KMT.</p>

<li>Which of the following is NOT an assumption of the KMT for a gas?</li>
</ol>

<p>(A) Gases are made up of tiny particles in cosntant chaotic motion.
(B) Gas particles are very small compared to the avg distance between the particles.
(C) Gas particles collide witht he walls of their container in elastic collisions.
(D) The avg velocity of the gas particles is directly proportional to the absolute temperature.
(E) All of the above are correct.</p>

<p>63.</p>

<p>PV=nRT
n=PV/RT = (1)(0.14)/(0.08206)(273) = 0.006 mol O2</p>

<p>0.006 mol O2 x (2 mol KClO3)/(3 mol O2) x (122.55 g KClO3)/1 mol KClO3) = 0.4902 g KClO3.</p>

<p>(0.4902 g KClO3)/(1.80 g) = 0.284 = 28.4%</p>

<p>67.</p>

<p>P<em>H2 = P</em>total - P_vapor = 0.810 atm - 0.024 atm = 0.786 atm</p>

<p>PV=nRT
n=PV/RT= (0.786)(1.7)/(0.08206)(293) = 0.056 mol H2.</p>

<p>72.</p>

<p>760 torr - 26 torr = 734 torr = 0.97 atm O2</p>

<p>PV=nRT
n=PV/RT = (0.97 atm O2)(2.00 L)/(0.08206)(300K) = 0.0788 mol O2</p>

<p>0.0788 mol O2 x (2 mol KClO3)/(3 mol O2) = 0.0525 mol KClO3</p>