HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq) If the value of Kc for this process is 5.2 x 10-5, there is no H3O+ or BrO- initially present, and the initial [HBrO] = 0.749 M, what will be the concentration of BrO- at equilibrium?

HINT: you should not use the quadratic formula for this problem.

7.67 x 10-3 M

9.01 x 10-3 M

5.17 x 10-3 M

4.29 x 10-3 M

6.24 x 10-3 M

6.24×10¯³ M

Explanation:

Step 1:

The equation for the reaction is given below:

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

Step 2:

The following data were obtained from the question:

Equilibrium constant, Kc = 5.2x10^-5

Concentration of HBrO, [HBrO] = 0.749 M

Concentration of BrO-, [BrO-] =?

Step 3:

Determination of the concentration of

BrO-.

We can obtain the concentration of BrO- as follow:

Initial concentration:

[HBrO] = 0.749 M

[H3O+] = 0

[BrO-] = 0

During reaction:

[HBrO] = - y

[H3O+] = +y

[BrO-] = +y

After reaction:

[HBrO] = 0.749 - y

[H3O+] = y

[BrO-] = y

Applying the equation constant equation:

Kc = [H3O+] [BrO-] / [HBrO]

5.2x10^-5 = y × y / 0.749

5.2x10^-5 = y² / 0.749

Cross multiply

y² = 5.2x10^-5 x 0.749

Take the square root of both side

y = √(5.2x10^-5 x 0.749)

y = 6.24×10¯³ M

[BrO-] = y = 6.24×10¯³ M

Therefore, the the concentration of BrO- at equilibrium is 6.24×10¯³ M

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.