Mallory combines two chemicals and notices that each chemical retains its original properties. What did Mallory form?

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

7 0

2 years ago

How many atoms are in 169g of calcium?

uysha [10]

The answer is 6.022• 10^23 atoms

5 0

3 years ago

Write the molecular formula of ferric oxide and calcium hydroxide by criss-cross method

irina [24]

<h3>Ferric Oxide:-</h3>

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Fe}\put(0,0){\bf -3}\put(4,3){\sf O}\put(4,0){\bf -2}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}$

$\\ \sf\longmapsto Fe_2O_3$

<h3>Calcium Hydroxide:-</h3>

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,3){\bf Ca}\put(0,0){\bf 2}\put(4,3){\sf OH}\put(4,0){\bf -1}\put(0.2,2.9){\vector(4,-3){3.5}}\put(3.8,2.9){\vector(-4,-3){3.5}}\end{picture}$