What is the Henry's law constant for Ryersonite dioxide in water at 25°C if (RyO2] =4.5 M in solution when the partial pressure of RyO2 above the solution is 0.065
atm?
0.047 M/atm
0.014 M/atm
0.031 M/atm
69 M/atm
45 M/atm

69 M/atm

Explanation:

From the question given above, the following data were obtained:

Concentration (C) = 4.5 M

Partial pressure (P) = 0.065 atm

Henry's law constant (K) =?

The Henry's law constant, K can be obtained as follow:

C = KP

4.5 = K × 0.065

Divide both side by 0.065

K = 4.5 / 0.065

K = 69 M/atm

Thus, the Henry's law constant for Ryersonite dioxide is 69 M/atm.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.