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4 months ago

Which change of state is shown in the model?

Chemistry

2 answers:

alisha [2.9K]3 months ago

7 0

$\boxed{\text{Deposition}}$ is depicted in the provided model.

Further Explanation:

Condensation

This is the process where a gas or vapor transitions to a liquid state. It is essentially the inverse of evaporation. The chaotic movement of gas particles decreases, allowing them to coalesce into a liquid. This transformation is influenced by changes in temperature and pressure.

Deposition

This refers to the phase change where a gas or vapor transforms directly into a solid without entering the liquid state. This process is thermodynamic in nature. Deposition occurs when water vapor loses sufficient thermal energy to become solid without going through the liquid phase. This process is the opposite of sublimation and is often referred to as desublimation.

Boiling

This phase change allows a liquid to turn into a gas or vapor state, which occurs when the liquid is heated to its boiling point.

Freezing

This conversion changes a substance from a liquid state to a solid state. It is the opposite of melting. In this phase transition, heat is released, causing the liquid particles to move closer together and form a solid, such as in the formation of ice.

In the model shown, the gaseous or vapor state is converted into a solid state, illustrating deposition. Hence, deposition is shown in the given model.

Learn more:

Identify the phase change in which crystal lattice is formed:
The main purpose of conducting experiments:

Answer details:

Grade: Senior School

Chapter: Phase transition

Subject: Chemistry

Keywords: deposition, freezing, boiling, solid, liquid, vapor, condensation, desublimation, thermodynamic process.

Alekssandra [3K]3 months ago

5 0

I think the state change illustrated in the diagram is deposition.
Deposition is the transformation of gases into solids without transitioning through a liquid phase. It is the reverse process of sublimation.
A key distinction between gases and solids lies in the spacing of molecules; gases have large spaces between molecules, whereas solids have very minimal spacing, resulting in solids being more densely packed. This is illustrated in the diagram showing the transition from gases to solids.

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Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculat

Tems11 [2777]

The question is not fully stated; here is the full version:

Using this data alongside the standard enthalpies of formation for $O_2(g)$ , $CO_2(g)$ , and $H_2O(l)$ found in Appendix C, determine the standard enthalpy of formation for acetone.

The complete combustion of 1 mole of acetone $(C_3H_6O)$ releases 1790 kJ:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ$

Answer: The standard enthalpy of formation for $CO_2(g)$ is calculated to be -247.9 kJ/mol

Explanation:

Enthalpy change represents the variation in enthalpy for all products and reactants based on their respective mole counts. This is denoted as $\Delta H^o$

The enthalpy change calculation for a chemical reaction follows this equation:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

Concerning the chemical reaction in question:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)$

The equation reflecting the enthalpy change for this reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})]$

Provided data includes:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ$

Substituting values from the equation gives us:

$-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol$

Thus, the enthalpy of formation of $C_3H_6O(g)$ computes to -247.9 kJ/mol.

4 0

3 months ago

A laser produces red light of wavelength 632.8 nm. Calculate the energy,

KiRa [2933]

Answer:

189.2 KJ

Explanation:

Provided Data

light wavelength = 632.8 nm

Convert nm to meters

1 nm = 1 x 10⁻⁹

632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m

What is the energy of 1 mole of photons?

Solution

Used Formula

E = hc/λ

where

E = energy per photon

h = Planck's Constant

Planck's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light

speed of light = 3 × 10⁸ ms⁻¹

λ = wavelength of light

Insert values into the equation

E = hc/λ

E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)

E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)

E = 3.141 x 10⁻¹⁹J

3.141 x 10⁻¹⁹J indicates the energy for a single photon

Next, we need to determine the energy for 1 mole of photons

It is known that

1 mole contains 6.022 x10²³ photons

Consequently,

Energy for one mole of photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³

Energy for one mole of photons = 1.89 x 10⁵ J

Now convert J to KJ

1000 J = 1 KJ

1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ

Thus,

the energy for one mole of photons is 189.2 KJ

3 0

2 months ago

Compound A reacts with Compound B to form only one product, Compound C, and it's known the usual percent yield of C in this reac

VMariaS [2998]

Response:

1. 16.54 grams.

2. 6.64 grams.

Clarification:

Greetings,

In this situation, the chemical reaction taking place is:

$A+B\rightarrow C$

Thus, based on the provided data, the theoretical yield of C can be determined as:

$m_C^{theoretical}=\frac{12.9g}{0.78}=16.54g$

Additionally, taking into account the principle of conservation of mass, the mass prior to the reaction is equal to the mass following the reaction, hence, the mass of B that was utilized amounts to:

$m_B=m_C-m_A=16.54g-10.0g=6.64g$

Best regards.

6 0

2 months ago