Ask question

Ask question

All categories

2 years ago

7

Describe the particles of a solid, liquid, and gas. For each, be sure to tell how far apart they are and how fast or slowly they

move.

1 answer:

Art [367]2 years ago

3 0

Answer:

Solid

A solid is a state of matter where the particles are tightly packed together, they are very close together

They vibrate, they move very slowly

Liquid

A liquid has free-flowing particles, but they may not leave the container
They move at medium speed

Gas

<u><em>A gas moves very freely, they leave the container, and they may not be captured in its entirety</em></u>
<u><em>It moves very fast because it is heated</em></u>

<u><em></em></u>

Hope this helps :)

- jp524

You might be interested in

Which describes the path of energy that a gas powered car uses to drive along a road?

harkovskaia [24]

Answer:

mechanical energy to electrical energy to light energy

4 0

3 years ago

Consider the reaction below. At 500 K, the reaction is at equilibrium with the following concentrations. [PCI5]= 0.0095 M [PCI3]

Dmitriy789 [7]

Answer: The equilibrium constant for the given reaction is 0.0421.

Explanation:

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Concentration of $[PCl_5]$ = 0.0095 M

Concentration of $[PCl_3]$ = 0.020 M

Concentration of $[Cl_2]$ = 0.020 M

The expression of the equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}$

$K_c=0.0421$ (An equilibrium constant is an unit less constant)

The equilibrium constant for the given reaction is 0.0421.

6 0

3 years ago

Read 2 more answers

Pls that is the picture

Fiesta28 [93]

Do what it says easy

3 0

2 years ago

andriy [413]

Answer:

1. true

2. false

3. true

Explanation:

5 0

2 years ago

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

<u>Part A:</u>

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

<u>Part B:</u>

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0

3 years ago