Homework #1 Solutions

Homework #1 Solutions - purdy (sbp456) – Homework 1 –...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: purdy (sbp456) – Homework 1 – flowers – (53885) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The formation of tiny bubbles when a beaker of water is mildly heated indicates that 1. water is being decomposed into hydrogen and oxygen. 2. the liquid is at its boiling point. 3. air is less soluble in water at higher tem- peratures. correct 4. water is being extensively hydrogen bonded. 5. the lattice energy of water is large. Explanation: The bubbles are tiny pockets of air that were dissolved in the water at the lower tem- perature. As the gas molecules acquired more energy they were able to escape from the sol- vent. 002 10.0 points Calculate the concentration of argon in lake water at 20 ◦ C. The partial pressure of argon is 0.0090 atm and Henry’s constant is 0.0015 mol · L − 1 · atm − 1 . 1. 6.0 M 2. 1.4 × 10 − 5 M correct 3. 9.0 × 10 − 3 M 4. 0.17 M 5. 1.5 × 10 − 3 M Explanation: 003 10.0 points Based on the types and strengths of in- termolecular forces present, place the com- pounds C 6 H 14 , C 3 H 8 , C 5 H 11 OH , C 5 H 11 Cl , CaBr 2 in order from lowest to highest boiling point (lowest on the left, highest on the right). 1. C 6 H 14 , C 3 H 8 , C 5 H 11 OH , C 5 H 11 Cl , CaBr 2 2. CaBr 2 , C 5 H 11 Cl , C 6 H 14 , C 5 H 11 OH , C 3 H 8 3. CaBr 2 , C 5 H 11 OH , C 5 H 11 Cl , C 6 H 14 , C 3 H 8 4. C 3 H 8 , C 6 H 14 , C 5 H 11 OH , C 5 H 11 Cl , CaBr 2 5. C 3 H 8 , C 5 H 11 OH , C 6 H 14 , C 5 H 11 Cl , CaBr 2 6. C 3 H 8 , C 6 H 14 , C 5 H 11 Cl , C 5 H 11 OH , CaBr 2 correct Explanation: C 3 H 8 and C 6 H 14 have only dispersion forces, the weakest type of intermolecular forces. C 6 H 14 has a higher molecular weight and will therefore have stronger dispersion forces. C 4 H 11 Cl and C 5 H 11 OH are both approx- imately the same size as C 6 H 14 and would therefore have about the same strength of dis- persion forces. C 4 H 11 Cl also has dipole-dipole interactions, stronger than dispersion forces. C 5 H 11 OH also has H -bonding, stronger than dipole-dipole interactions. CaBr 2 is ionic and has interionic forces, the strongest type of intermolecular forces. The stronger the intermolecular forces, the higher the boiling point. 004 10.0 points Which of the following molecules A) CH 3 OCH 3 C) CH 3 CH 2 OH B) CH 3 COOH D) CH 3 CHO are likely to form hydrogen bonds? 1. C and D only purdy (sbp456) – Homework 1 – flowers – (53885) 2 2. B and C only correct 3. None forms hydrogen bonds. 4. A and C only 5. Another combination of compounds 6. A and D only 7. B and D only 8. A and B only 9. All are likely to form hydrogen bonds....
View Full Document

This note was uploaded on 11/06/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

Page1 / 8

Homework #1 Solutions - purdy (sbp456) – Homework 1 –...

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