All categories

3 years ago

what if you drop a small magnet repeatedly it become demagnetized. explain. what is happening sub-atomically?

Chemistry

1 answer:

xenn [34]3 years ago

3 0

I guess the the alignments of the atoms in the magnet are being demagnetized when it's dropped

You might be interested in

Identify all the examples that support the Law of Conservation of Mass through chemical changes.

wariber [46]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

How many meters are their in one kilometer

tatuchka [14]

There are 1000 meters in 1 Kilometer.

Hope this helps you. :)

5 0

3 years ago

Read 2 more answers

6) Which country is located south of New Mexico and Arizona?

Dvinal [7]

Answer:

mexico

Explanation:

3 0

3 years ago

Please please !!! help me I’ve been doing this for 2 hours

MatroZZZ [7]

Answer:

3.7 x 10^23

is the right answer

6 0

2 years ago

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