Physic0 -Chemical Properties
and
Environmental
Fate of Pesticides
BY
Clark D. Jinde
Student Intern, U.C. Davis
January
1994
ENVIRONMENTAL HAZARDS ASSESSMENT PROGRAM
STATE OF CALIFORNIA
Environmental Protection Agency
Department of Pesticide Reguiation
Env
Assignment 8 Solutions
1. An adsorption study was conducted by adding varying amounts of
activated carbon to a series of seven flasks containing 500mL of feed
water used in soft drink preparation having an initial TOC of 2.0mg/L.
The flasks were agitated
What should you know?
Soil colloids - what they are, their
properties
Differences between soil clay minerals
Properties of humic substances (active
organic matter)
Cation exchange and CEC
Soil Colloids and
Cation Exchange Capacity
Chapter 8
p. 316-362
Water-Quality Principles
QW1022TEL
Lesson 8Environmental Organic Chemistry
Source: Girard, J.E., 2010, Principles of environmental chemistry (2d ed.): Sudbury, Mass.,
Jones and Bartlett Publishers, p. 413418 (http:/www.jblearning.com/). Used with
permis
APPENDIX C
WORK PLAN FOR DISTRIBUTION COEFFICIENT COLUMN STUDY
Introduction
The primary objective of this study is to measure the distribution coefficient (Kd) for perchlorate in the
soil samples that are scheduled to be collected from the borings compl
Organic SolventWater Partitioning; Bioconcentration D 1
D. ORGANIC SOLVENT-WATER
PARTITIONING; BIOCONCENTRATION
ILLUSTRATIVE EXAMPLES
Estimating the Activity Coefficients of Organic
Compounds in Organic Phases
Problem
Calculate the activity coefficients
Stat 351 Homework #5. Chapter 16.1
Due March 28th Tuesday by 9:20 am.
Problem 1: exercise 3 in Chapter 16, part a, b, c, d.
Problem 2:
a). What is the form of the general linear model? The linear model is called linear in
terms of what parameters?
b). In
Homework on regression.
1. Read Sections 5.1, 5.2, and 5.8.
2. Work problem 5.1 parts a, b, c and problem 5.16 in apply your knowledge.
Hand in the following problem next time.
Child development researchers explored the possible relationship between an in
Solutions to Homework 13 Problem 13.5 First, SSE = 1800 1200 = 600. The degrees of freedom for the Treatments is k 1 = 3 1 = 2. The degrees of freedom for the Error is nT k = 47 3 = 44. The degrees of freedom for the Total is nT 1 = 47 1 = 46. We then hav
Solutions to Homework 12 Problem 10.5 a. The point estimate is x1 x2 = 135.67 68.64 = 67.03. b. The margin of error is M OE = z/2 c. The 99% condence interval is (1 x2 ) M OE = 67.03 17.08 = (49.95, 84.11). x Problem 10.8 a. The test statistic is z= (1 x2
Solutions to Homework 11 Problem 9.25 For all parts of the problem 0 = 45 and n = 36. This means we have n 1 = 36 1 = 35 degrees of freedom for our t distribution. a. The test statistic is t= 44 45 x 0 = 1.15. = s/ n 5.2/ 36
The p-value is between 0.10 an
Solutions to Homework 10 Problem 9.1 a. The hypotheses are H0 : 600 Ha : > 600 The alternative hypothesis should correspond to the claim that the bills have been increasing ( > 600). b. If H0 cannot be rejected then we would conclude that it is possible t
Solutions to Homework 9 Problem 8.29 a. We can compute n:
2 z/2 2 1.962 6.252 = = 37.5. E2 22 So we should sample at least 38.
n=
b. We can compute n:
2 z/2 2 1.962 6.252 n= = = 150.06. E2 12 So we should sample at least 151.
Problem 8.35 a. The point est
Solutions to Homework 8 Problem 8.3 a. The 95% condence interval for the population mean is 15 x z/2 = 80 1.96 n 60 = 80 3.80 = (76.20, 83.80) b. If there were 120 items in the sample the 95% condence interval is 15 = 80 1.96 x z/2 n 120 = 80 2.68 = (77.3
Solutions to Homework 7 Problem 7.13 a. A point estimate of the population mean number of units sold per month is x= xi 94 + 100 + 85 + 94 + 92 = = 93. n 5
b. A point estimate of the population standard deviation is s= Problem 7.17 a. The total number of
Solutions to Homework 6 Problem 6.11 a. P (z 1.0) = 0.1587 b. P (z 1) = 1 0.1587 = 0.8413 c. P (z 1.5) = 1 0.0668 = 0.9332 d. P (2.5 z ) = 1 0.0062 = 0.9938 e. P (3 < z 0) = 0.5000 .0013 = 0.4987 Problem 6.13 a. P (1.98 z 0.49) = 0.6879 0.0239 = 0.6640 b.
Solutions to Homework 5 Problem 4.4 a. The tree diagram:
b. The possible experimental outcomes are HHH,HHT,HTH,HTT,THH,THT,TTH,TTT c. Each outcome is equally likely, so the probability for each outcome is Problem 4.9 Here we are counting the number of pos
Solutions to Homework 4 Problem 3.17 a. The mean price for models with a DVD player is 450 + 300 + 400 + 500 + 400 = 410. 5 The mean price for models without a DVD player is 300 + 300 + 360 + 290 + 300 = 310. 5 So it is about $100 additional to perchase a
Solutions to Homework 3 Problem 3.9 a. The mean number of minutes is xi 615 + 135 + + 105 6330 = = = 422 minutes. n 15 15 b. To compute the median we rst put the day in ascending order. 105 135 395 420 180 430 210 615 245 690 250 830 265 1180 380
We have
Solutions to Homework 2 Problem 2.9 a. and b. The frequency and percent frequency distributions are in the table below. Became CEO Inherited Built Hired c. The bar graph: Freq. 8 14 4 Relative Freq. 0.308 0.538 0.154 Percent Freq. 30.8% 53.8% 15.4%
d. 30.
Solutions to Homework 1 Problem 1.5 a. There are 5 variables in this data set. b. The variables Sound Quality and FM Tuning are qualitative. Price, CD Capacity, and Tape Decks are quantitative. c. The average CD capacity is 3. d. 7 minisystems have FM tun
Review and Examples Between-Treatments Estimate of Variance
Within-Treatments Estimate of Variance
The F Test
The ANOVA Table
Fishers LSD Procedure
Example: Job satisfaction was measured for a sample of 10 individuals from each of the following profes
Small
90.5 78.2 92.3 95.7 94.1 100 91.8 95
Medium Large
91.1 98.9 94.2 84.3 84.8 89.2 86.4 88.3 89.2 90.2 85.9 84.2 90.2 80.6 75.8 82.3
This file cont ains passenger r at ings for t hr ee differ ent sizes of c medium, and lar ge (higher r at ings mean hig
Review and Examples Hypothesis Tests for 1 2 consist of four steps: 1. State the hypotheses. The hypotheses should take one of these forms: H0 : 1 2 D0 Ha : 1 2 < D0 H0 : 1 2 D0 Ha : 1 2 > D0 H0 : 1 2 = D0 Ha : 1 2 = D0
2. State the level of signicance. T
College High School 485 442 534 580 650 479 554 486 550 528 572 524 497 492 592 478 487 425 533 485 526 390 410 535 515 578 448 469
The College Board provided comparisons of SAT scores based on the highest level of education attained by the test taker's p
Review and Examples Hypothesis Tests for means (with unknown ) consist of four steps: 1. State the hypotheses. The hypotheses should take one of these forms: H0 : 0 Ha : < 0 H0 : 0 Ha : > 0 H0 : = 0 Ha : = 0
2. State the level of signicance. This is denot
Review and Examples The hypotheses for a hypothesis test take one of the following forms: H0 : 0 Ha : > 0 H0 : 0 Ha : < 0 H0 : = 0 Ha : = 0
The form chosen depends on the problem being asked. 0 is just a number. The hypotheses are statements about the pa