Table 2. Soil moisture content data Sample Moisture state 1 Field capacity (2 days after rain) Wilting point (10 days after rain) Calculate the following quantities. Calculations from Sample 1 data are shown as an example. 2 Moist weight 160 g 170 g % water- (160g-128g) x 100% = 25% 128 g Oven-dry weight 128 g 156 g 1. What is the percent water by weight (gravimetric water content) in each sample? Sample 1 Sample 2 2. What is the percent water by volume (volumetric water content) in each sample? Sample 1 Sample 2 % water = 25% x 1.3 g/cm³ =33%
Table 2. Soil moisture content data Sample Moisture state 1 Field capacity (2 days after rain) Wilting point (10 days after rain) Calculate the following quantities. Calculations from Sample 1 data are shown as an example. 2 Moist weight 160 g 170 g % water- (160g-128g) x 100% = 25% 128 g Oven-dry weight 128 g 156 g 1. What is the percent water by weight (gravimetric water content) in each sample? Sample 1 Sample 2 2. What is the percent water by volume (volumetric water content) in each sample? Sample 1 Sample 2 % water = 25% x 1.3 g/cm³ =33%
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Could you help me with question one please?
Transcribed Image Text:Table 2. Soil moisture content data
Sample
1
2
Moisture state
Field capacity (2 days after rain)
Wilting point (10 days after rain)
Moist weight
160 g
170 g
Oven-dry weight
128 g
156 g
Calculate the following quantities. Calculations from Sample 1 data are shown as an example.
1. What is the percent water by weight (gravimetric water content) in each sample?
Sample 2
Sample 1
% water = (160g-128g) x 100% = 25%
128 g
2. What is the percent water by volume (volumetric water content) in each sample?
Sample 1
Sample 2
% water = 25% x 1.3 g/cm³ = 33%
Expert Solution
Step 1: Heating of a moist substance
Answer-1
When a moist substance is heated, it loses the water present in it and becomes dry and mass of water lost is equal to the difference between mass of moist substance and mass of dry substance.
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