Identify the parts of an experiment in the scenario below.

What is meant by conversion of energy

MissTica

the process of changing one form of energy into another

4 0

3 years ago

What is the name of the gas law relating pressure and volume at a constant temperature?

pochemuha

Answer:

Charles Law

Explanation:

Charles's law (also known as the law of volumes) is an experimental gas law that describes how gases tend to expand when heated. A modern statement of Charles's law is: This relationship of direct proportion can be written as: V∝T

5 0

3 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

3 years ago

Which has more particles, a mole of copper (Cu) atoms or a mole of neon

goldfiish [28.3K]

Answer:

D

Explanation:

it is neon because neon has a higher atomic number so it would have more protons and neutrons and electrons in one atom thus having more particles in one mole

7 0

3 years ago

Read 2 more answers

What fraction of a Sr-90 sample remains unchanged after 87.3 years

jolli1 [7]

The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
 $t_{1/2}$ = 28.8 years

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{87.3 years}{28.8 years}$
⇒ $n=3.03$
⇒ n ≈ 3

Now we can use the first equation to calculate the remained amount of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^3$
⇒ $x= \frac{1}{8}$

8 0

3 years ago