Calculate the molar solubility of Ag Cros in water. Use 1.10 x 10-12 as the solubility product constant of Ag2CrO4. O 7.42 x 10-7M
O 1.05 x 10-6 M
O 6.50 x 10-5 M
O 1.03 x 10-4 M

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

diaria maybe go see a doctor about that