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2 years ago

Which change of state is shown in the model?

Chemistry

2 answers:

nata0808 [166]2 years ago

7 0

$\boxed{\text{Deposition}}$ is shown in the given model.

Further Explanation:

Condensation

The process by which a substance in its gaseous or vapor form is transformed into its liquid form is termed as condensation. It is just the reverse of evaporation. The random motion of gas particles is reduced and they come together to form a liquid. This is done by altering the temperature and pressure of the substance.

Deposition

The phase transition due to which a gas or vapor transforms directly into solid by not going through the liquid phase is termed as deposition. It is a thermodynamic process. This process occurs when water vapors release enough of its thermal energy and get converted into solids directly without passing through the liquid state. It is the opposite of what is done in the sublimation process and is therefore known as desublimation.

Boiling

The phase transition due to which gaseous or vapor state is formed from the liquid state is known as boiling. This takes place when a substance in its liquid state is heated to its boiling point.

Freezing

It is a process in which the substance gets converted from its liquid state to a solid state. It is just the reverse of melting. In this phase transition, heat is released from the substance and the liquid particles come closer to form solid. The formation of ice is an example of freezing.

In the given model, gaseous or vapor state is transformed into a solid state and this phase change occurs in case of deposition. Therefore deposition is shown in the given model.

Learn more:

Identify the phase change in which crystal lattice is formed: brainly.com/question/1503216
The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: Senior School

Chapter: Phase transition

Subject: Chemistry

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

Romashka [77]2 years ago

5 0

I believe the change of state shown in the model is deposition.
Deposition is a process in which gases change phase and turns directly in solids without passing through the liquid phase. It is the opposite of sublimation.
One of the major difference between gases and solids is the distance between molecules; in gases the inter molecular spaces are large, while in solid they are very small, making solids be the most dense, with closely packed molecules. This is evident in the diagram, the phase changed from gases to solids.

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago

Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

Molar mass of solute $PO_4^{3-}$ = 94.97 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L$

Therefore, the normality of the solution is, 30.006 N

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We first calculate for the number of moles of NaOH by dividing the given mass by the molar mass of NaOH which is equal to 40 g/mol. Solving,
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Then, we divide the calculate number of moles by the volume in liters.
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molarity = 1.95 M

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