The following reaction has a value of kc = 3.07 x 10-4 at 25o c. a reaction is studied with the initial concentrations of [no] = 1.58 x 10-2 m, [br2] = 1.29 x 10-2 m, and [nobr] = 7.13 x 10-2 m
2 nobr (g) < > 2 no (g) + br2 (g)choose the one correct statement from the following: (a) qc = 1.58 x 104 and the concentration of nobr will decrease.(b) qc = 6.33 x 10-4 and the concentration of nobr will increase.(c) qc = 2.85 x 10-3 and the concentration of nobr will increase.
(d) qc = 2.85 x 10-3 and the concentration of nobr will decrease.

Qc = 6.33 x 10-4 and the concentration of NOBr will increase

Explanation:

2 NOBr (g) →  2NO (g) + Br₂(g)

Qc = Reaction quotient. It tells us the quantity of products and reactants during the reaction at a particular point in time, other than equilibrium.

The expression for Qc is the same for Kc

Qc = ([NO]² . [Br]) / [NOBr]²

Qc = ((1.58x10⁻²)² . 1.29x10⁻²) / (7.13x10⁻²)²

Qc = 6.33x10⁻⁴

You have to compare Qc to Kc to analyse the reaction:

Qc < Kc → Some reagents have not generated products yet.

Reaction tends to Products

Qc > Kc → Reaction tends to Reactants.

Reactants concentration must be increased to make more products

Qc = Kc →  Equilibrium

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

the answer is na (sodium)