HELP ME PLS Asap :Which type of line is shown on the map?

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

3 years ago

What is the name of this element?

Strike441 [17]

Answer:

Tungsten

Explanation:

8 0

3 years ago

A metal crystallizes in a body-centered cubic unit cell. The radius of one atom = 2.30 x 10 -8 cm. The density of the metal is 0

maw [93]

Answer:

25.41 g/mol is molar mass of metal.

Explanation:

Number of atom in BCC unit cell = Z = 2

Density of metal = $0.867 g/cm^3$

Edge length of cubic unit cell= a = ?

Radius of the atom of metal = r = $2.30\times 10^{-8} cm$

$a=2\times r=2\times 2.30\times 10^{-8} cm=4.60\times 10^{-8} cm$

Atomic mass of metal =M

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$0.867 g/cm^3=\frac{2\times M}{6.022\times 10^{23} mol^{-1}\times (4.60\times 10^{-8} cm)^{3}}$

$M = 25.41 g/mol$

25.41 g/mol is molar mass of metal.

6 0

4 years ago

Which option lists the layers of the rainforest in the correct order from top to bottom?

andrey2020 [161]

Answer: B.

Explanation:

Trust me already took the quiz on it its B.

8 0

3 years ago

Read 2 more answers

Identify the nuclide that would be formed from the alpha particle emission of Fm-257.

Elodia [21]

Answer: 1) The nuclide formed from the alpha decay of $_{100}^{257}\textrm{Fm}$ is $_{98}^{253}\textrm{Cf}$

2) The nuclide formed from the beta-minus decay of $_{98}^{253}\textrm{Cf}$ is $_{99}^{253}\textrm{Es}$

Explanation:

1) For alpha-decay

In this process, alpha particle is released.

Equation follows:

$_{100}^{257}\textrm{Fm}\rightarrow _{98}^{253}\textrm{Cf}+_2^4\alpha$

The nuclide formed is $_{98}^{253}\textrm{Cf}$

2) For beta-minus decay

In this process, a neutron is converted into a proton and an electron.

Equation follows:

$_{98}^{253}\textrm{Cf}\rightarrow _{99}^{253}\textrm{Es}+_{-1}^0\beta$

The nuclide formed is $_{99}^{253}\textrm{Es}$

5 0

3 years ago