All categories

3 years ago

Why does changing the volume of a container change the pressure of the gas in it?

Chemistry

2 answers:

ollegr [7]3 years ago

8 0

Answer:

Changing the volume increases the area that the molecules collide with so the force is spread over a larger area.

Explanation:

The volume of a container is the space within the container. This is the amount of void the gas can occupy.

When the volume of a container is changed, the space either increases or decreases.

An increase in volume creates more space within the container.
This spreads out the force over a larger area.

When the volume of the container is reduced,

the force is confined to much smaller space.

The frequency of collision is higher for tiny spaces.

But lower for a large volume.

Gemiola [76]3 years ago

5 0

Answer:

D on EDG

Explanation:

You might be interested in

Consider the reaction.

Stella [2.4K]

Answer:

when the rates of the forward and reverse reactions are equal

Explanation:

In a chemical system, the reaction reaches a dynamic equilibrium when the rate of formation of product equals the rate of formation of reactants. This implies that both the forward and revered(backwards) reaction are occurring at the same rate.

5 0

3 years ago

What term is defined as splitting large nuclei into smaller ones

svetoff [14.1K]

Answer:

Nuclear Fission.

Explanation:

This happens when a high-energy particle collides with a radioisotope, which splits into 2 daughter nuclei, several neutrons (which can collide with more radioisotopes to cause a chain reaction); and a lot of energy. That's why nuclear power plants are so good.

5 0

3 years ago

How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?

lara [203]

<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

<em>q</em> is heat (in J)
<em>m</em> is mass (in g)
<em>c</em> is specific heat (in J/g °C)
ΔT is change in temperature (in °C or K)

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
Multiply: q = (941.4 J/°C)(159.8 °C)
Multiply: q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0

3 years ago

N204(0) + 2NO2(g)

user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

$\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}$

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state

Q is the product of the ion ions from the reacting substance

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

$\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}$