Ineed with a, b, c, d, and e on the long answer

Answers:

(a) Second order in A

(b) Second order in B

(c)     r = k[A]²[B]²

(d)    k = 0.0123 L³mol⁻³s⁻¹

(e) [A] = 8.86 mol·L⁻¹

Step-by-step explanation:

A + B ⟶ C

The rate law is r = k [A]^m[b]ⁿ

Our problem is to determine the values of m and n.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] doesn't.

They would be Experiments 3 and 1.

[B] is constant, so only [A] is changing the rate

r₃/r₁ = {k[A]₃^m}/{k[A]₁^m }     Cancel the ks

r₃/r₁ = {[A]₃/[A]₁}^m

80/5 = (40.2/10.1)^m

   16 = 3.98^m

(i) Method 1: by inspection

16 ≈4²

If quadrupling the concentration multiplies the rate by a factor of 16 (4²), the reaction is 2nd order.

(ii) Method 2: mathematical

    16 = 3.98^m

log16 = mlog3.98     Take the logarithm of each side

    m = (log16)/(log3.98)

       = 1.204/0.5999

       = 2.007             Round off to nearest integer

       ≈ 2

By either method, the reaction is second order in A.

r = k[A]²

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] doesn't. There are none.

However, we know the effect of A on the rate.

Choose a different pair, say, Experiments 2 and 1.

       r₂/r₁ = ([A₂]/[A]₁)²([B]₂/[B₁]^m

     80/5 = (19.8/10.1)²(3.99/2.01)^m

         16 = (1.960)²(1.985)ⁿ

         16 = 3.843 × 1.985ⁿ     Divide each side by 3.843

    4.163 = 1.85ⁿ                     Take the logarithm of each side

log4.163 = nlog1.985            Divide each side by log1.985

           n = log4.163/log1.985

              = 2.08

              ≈ 2

The reaction is second order in B.  

rate = k[B]²

(c) Overall rate law

The reaction is second order in A and second order in B.

The overall rate law for the reaction is

r = k[A]²[B]²

(d) Value of k

Choose any experiment (say, Experiment 3) and insert the known values.

                    r = k[A]²[B]²

80 mol·L⁻¹·s⁻¹ = k(40.2 mol·L⁻¹)²(2.00 mol·L⁻¹)²

                      = k × 1616 mol²L⁻² × 4.00 mol²L⁻²

                      = k × 6464 mol⁴L⁻⁴     Divide each side by 6464 mol⁴L⁻⁴

                  k = (80 mol·L⁻¹s⁻¹)/(6464 mol⁴L⁻⁴)

                     = 0.0123 L³mol⁻³s⁻¹

(e) [A] in Experiment 4

    r = k[A]²[B]²

35 =  0.0123 × [A]²[6.00]²

      = 0.4456[A]²     Divide each side by 0.4456

[A]² = 35/0.4456

      = 78.6               Take the square root of each side

[A] = 8.86 mol·L⁻¹

well, they made a movie about how animals lived there.

rotten tomato score 95%

note: mark other user brainliest.!

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.