Please help! Answer please!

create visual representations that show the interactions between the particles in both samples during vaporization

ivolga24 [154]

Given what we know, we can confirm that despite not being able to provide a visual representation, if you were to create one, it would show the atoms in the substance moving with more and more energy over time until vaporization occurred.

<h3>What is vaporization?</h3>

Vaporization is the changing of a liquid into a gas.
This is known as a phase change.
This happens when the atoms receive enough heat energy.
The heat energy provides the atoms with kinetic energy and causes them to move faster over time.
When the kinetic energy is enough to rupture the bonds that hold a liquid together, it becomes a gas.

Therefore, given that vaporization is when atoms within a substance gain so much kinetic energy that they are able to rupture the bonds that hold them relatively close together and change into a gas, we can confirm that a visual representation would have to be a diagram showing the increased movement of atoms over time.

To learn more about vaporization visit:

brainly.com/question/11782435?referrer=searchResults

7 0

2 years ago

What statement correctly describes gravity

Sedbober [7]

Answer:

B

Explanation:

3 0

3 years ago

Read 2 more answers

A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and

pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

3 0

2 years ago

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

Answer:

For 1: The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2: The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

Explanation:

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

For 1:

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2:

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

Read 2 more answers

We participate in the ______________________cycle every time we breathe. tectonic nitrogen phosphorus carbon-oxygen

podryga [215]

Carbon-Oxygen
We breathe in oxygen, we breathe out carbon dioxide.

8 0

3 years ago