. The determination in problem 15-9 was modified to use the standard addition method. In this case, a 4.236 g tablet was dissolved in sufficient 0.10M HCl to give 1.00 L. Dilution of 20.00 mL aliquot to 100 mL yielded a solution that gave reading of 448 at 347.5 nm. A second 20.00 mL aliquot was mixed with 10 mL of 50 ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 525. Calculate the percentage of quinine in the tablet.

Imagine that a new planet is discovered with two moons of equal mass: moon a and moon b. the mass of the new planet is greater than the combined mass of its moons. moon a is farther away from the new planet than moon b. what is the planet's gravitational pull on moon a compared to the planet's gravitational pull on moon b? the planet's gravity repels moon a with a greater force than it repels moon b, which is why moon a is farther away. the gravitational pull on moon b is greater than on moon a because moon b is closer to the new planet than moon a. the gravitational pull on moon b is greater than on moon a because moon b is farther away from the new planet than moon a. the gravitational pull on moon a is the same as the gravitational pull on moon b because distance does not affect the planet's gravity.