Over an interval of 3.00 s the average rate of change of the concentration of C was measured to be 0.0840 M/s. What is the final concentration of D at the end of this same interval if its concentration was initially 3.000 M?

Final Concentration = 3.252M

Explanation:

Let's bring out the data mentioned in the question.

Time, t = 3s

Rate = 0.0840 M/s

Initial Concentration = 3M

Final Concentration = x

The formular relating these parameters is;

Rate = ΔConcentration / ΔTime

Upon solving for ΔConcentration, we have;

ΔConcentration = Rate * ΔTime

ΔConcentration = 0.0840 * 3

ΔConcentration = 0.252

But ΔConcentration = Final Concentration - Initial Concentration

Hence,

Final Concentration = ΔConcentration + Initial Concentration

Final Concentration = 0.252 + 3

Final Concentration = 3.252M

gas

explanation:

smoke is the matter of a gas