What is the concentration of silver ions where silver iodide, Agl, is in a solution of hydroiodic acid, HI, that has a pH of 3.55? Ksp = 8.51x10-17

Explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

Now, we can set up the equilibrium expression as shown below:

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

Best regards!

the correct answer is boyle

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.