What is the rate law for the reaction below: a + b + c dif the following data were collected? exp [a]0 [b]0 [c]0 initial rate1 0.4 1.2 0.7 2.32x10-32 1.3 1.2 0.9 7.54x10-33 0.4 4.1 0.8 9.25x10-24 1.3 1.2 0.2 7.54x10-3this is the explanation i was given but i still don't understan it. can you show the steps for solving the exponents of the rate laws "with algebra"? : the general form of the rate law has to be: rate = k [a]x [b]y [c]zusing the experiments given to set up several equations, we get: 2.32x10-3 = k(0.4)x(1.2)y(0.7)z7.54x10-3 = k(1.3)x(1.2)y(0.9)z9.25x10-2 = k(0.4)x(4.1)y(0.8)z7.54x10-3 = k(1.3)x(1.2)y(0.2)zwith algebra, we find x=1, y=3, and z=0. plugging these values in, we get the rate law as: rate = k [a] [b]3 [c]0 = k [a] [b]3we can now plug in any experiment into the rate law above to solve for k. plugging in experiment 1: 2.32x10-3 = k(0.4)(1.2)3k = 3.36x10-3the complete rate law: rate = 3.36x10-3 [a] [b]3

rate = k  [A] [B] ³

Explanation:

Set up a table with the data given in the question to study the dependence of the reaction rate on the concentrations of reactants.

[A]             [B]             [C]             Rate           Experiment #

0.4            1.2             0.7          2.32 x 10⁻³        (1)

1.3            1.2             0.9          7.54 x 10⁻³        (2)

0.4             4.1            0.8          9.25 x 10⁻²        (3)

1.3              1.2             0.2         7.54 x 10⁻³        (4)

The rate law will have the form : rate =  k [A] ^x   [B]  ^y    [C]   ^z

Comparing experiments (2) and (4) we have to conclude that  the rate  is zero order with respect to  [C]  since keeping [A]  and [B]  the same and varying [C] did not change the rate (i.e ,no dependence on  [C]) .

Now we know the rate law has the form rate =  k [A] ^x  [B]  ^y

Comparing (1) and (4) we keep [B] constant and increase [A] by a factor of  1.3/.4 = 3.25 and the rate increased by a factor of 0.00754 / 0.00232 =3.25, so we can conclude that the rate law is first order with respect to  [A]

Finally, comparing (1) and (3) while keeping  [A]  constant  increasing [B] by a factor of 4.1/1.2 = 3.416, the rate increased by a factor of 0.0925/0.00232 = 40, it is not entirely clear the dependence with respect to  [B] .

In this case we can always set up the following equation which is obtained by dividing  equation (3) by (1)

(4.1 / 1.2)^x = 0.0925/0.00232

taking natural log to both sides of the equation

x ln 3.4167 = ln 40

x = 3.69/1.23 = 3

So the dependence with respect to   [C] is three.

The rate law is :

rate = k  [A] [B] ³

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.