Common use of Illustrative Example Clause in Contracts

Illustrative Example. (i) Chemical A was dissolved in 0.010 M pH 7.0 buffer. The solution was filtered through a 0.2 )m filter, air saturated, and analyzed. It contained 1.7×10 ¥5 M A, five-fold less than its water solubility of 8.5×10 ¥5 M at 25 °C. A uv spectrum (1-cm path length) versus buffer blank showed no absorbance greater than 0.05 in the wavelength interval 290 to 825 nm, a condition required for the Phase 2 pro- tocol. The 180 mL mixture was diluted by the addition of 20 mL of SHW stock solution. (ii) The SHW solution of A was photolyzed in sealed quartz tubes (12×100 mm) in the fall season starting on October 1. At the end of 1 and 2 days, respectively, the concentration of A was found to be 1.13×10 ¥5 M and 0.92×10 ¥5 M compared to unchanged dark controls (1.53×10 ¥5 M). (iii) The tube photolysis rate con- stant of chemical A was calculated from Equation 2 under paragraph (c)(2)(vi)(B) of this section. The first time point at day 1 was used because the fraction of A remaining was in the range 20 to 80 percent:

Illustrative Example. (i) Chemical A was dissolved in 0.010 M pH 7.0 buffer. The solution was filtered through a 0.2 )m filter, air saturated, and analyzed. It contained 1.7×10 1.7 × 10 ¥5 M A, five-fold less than its water solubility of 8.5×10 8.5 × 10 ¥5 M at 25 °C. A uv spectrum (1-cm path length) versus buffer blank showed no absorbance greater than 0.05 in the wavelength interval 290 to 825 nm, a condition required for the Phase 2 pro- tocolprotocol. The 180 mL mixture was diluted di- luted by the addition of 20 mL of SHW stock solution. (ii) The SHW solution of A was photolyzed in sealed quartz tubes (12×100 12 × 100 mm) in the fall season starting on October 1. At the end of 1 and 2 days, respectively, the concentration of A was found to be 1.13×10 1.13 × 10 ¥5 M and 0.92×10 0.92 × 10 ¥5 M compared to unchanged dark controls (1.53×10 1.53 × 10 ¥5 M). (iii) The tube photolysis rate con- stant of chemical A was calculated from Equation 2 under paragraph (c)(2)(vi)(B) of this section. The first time point at day 1 was used because the fraction of A remaining was in the range 20 to 80 percent: