Three Rivers Community College

ENV 2101 Principles of Environmental Engineering I

Instructor: Anthony Benoit

Office: Room 201B (860) 885-2386

Fall ‘01 (Th 6:00-8:45 pm)

abenoit@trcc.commnet.edu

http://environmentalet.org/env2101

Problem

Answer

2.1

(b) 5 mol O₂ = 1 mol C₃H₈
(c) 363.2 g O₂
(d) 0.255 m³ O₂ and 1.213 m³ air
(e) 0.153 m³ CO₂

2.15

pH = 3.16

2.18

(a) 213 mg/L
(b) 62.6 mg/L
(c) Note: Do calculation for 50 mg/L of glucose, C₆H₁₂O₆
Answer: 53.3 mg/L

2.20

at pH 6: 0.92; at pH 8: 0.104

2.21

9.5*10^-7 mg/L

2.23

2.9 g/L; 0.067 M

2.24

pH = 3.76