HI decomposes into its elements according to second-order kinetics. What concentration of HI remains from an initial concentration of 2.30 M after 4.5 hours? The rate constant, k, equals 1.6 x 10-3 M-1 hr-1.

Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.

Step-by-step explanation:

second order reaction: The rate of reaction  proportional to the square of the concentration of reactant.

For second order reaction

K is rate constant = 1.6×10⁻³M⁻¹hr⁻¹

a = initial concentration of reactant = 2.30 M

a-x = concentration of reactant after t h

t = time = 4.5 h

Putting the values in the above equation,

Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.

nope

step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq