Problem Set 11 Key CHEM103A Dr. Keller

Problem Set 11 Key CHEM103A Dr. Keller - CHEMISTRY 103A...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEMISTRY 103A, Sections 7-10 Problem Set 11: 20 Points Answer Key The first four problems deal with the phase diagram below: 1. At which point are three phases coexisting in equilibrium? (a) G (b) D (c) B (d) H (e) C This is C, the triple point 2. What phase(s) exist in equilibrium at point H? (a) solid and liquid (b) liquid only (c) solid and gas (d) liquid and gas (e) supercritical fluid only
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
H is on the liquid-gas equilibrium line 3. In this diagram the solid-liquid line implies (a) the solid melts below -90 o C (b) The solid is denser than the liquid (c) The liquid is denser than the solid (d) The solid does not melt above -30 o C (e) Solid does not sublime. The denser form of the substance is always favored at higher pressure. Since the S/L line slopes right, the liquid will become solid at high enough pressure without changing the temperature. 4. At which point do the liquid and gas phases have the same density? (a) G (b) D (c) B (d) H (e) C This is the critical point. The end of the L/G line. 5. A sample of methane gas is compressed at a constant temperature until the pressure is quadrupled (x4). If the volume occupied by the gas before compression was 12.0 L, what is the final volume? (a) 48.0 L (b) 24.0 L (c) 6.0 L (d) 3.0 L (e) 16.0 L From PV = nRT if n and T do not change, the product of P and V is constant (Boyle's Law). If P quadruples, V must shrink to . 6. The balanced equation below represents the reaction of nitrogen with oxygen to form dinitrogen tetroxide: N 2 (g) + 2 O 2 (g) → N 2 O 4 (g) If 14.0 g of nitrogen is combined with 28.0 g of oxygen, which substance is the limiting reactant? (a) both are limiting (b) N 2 (c) N 2 O 4 (d) O 2 (e) no limiting reactant Looking at the balanced equation, one can see that 1 mole (28 g) of nitrogen requires 2 mol (64 g) of oxygen; more than twice as much in terms of weight. Since the given weight of oxygen (28
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/14/2008 for the course CHEM 103a taught by Professor Weso during the Fall '08 term at Arizona.

Page1 / 6

Problem Set 11 Key CHEM103A Dr. Keller - CHEMISTRY 103A...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online