Consider a reactant with an order of 1. What effect does that reactant have on the rate equation when the concentration is doubled?

The rate of the reaction gets doubled when the concentration of the reactant is doubled.

Explanation:

We are given:

Order of the reaction = 1

Rate law equation for any order follows:

where, n = order of the reaction

= concentration of the reactant A

Rate law equation for the first order reaction becomes:

      .........(1)

When the concentration of the reactant is doubled, the new rate law becomes:

    .....(2)

Taking the ratio of equation 2 and equation 1, we get:

Hence, the rate of the reaction gets doubled when the concentration of the reactant is doubled.

12 is it do you think so

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally