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3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :) How is molecular motion related to phase changes?

2 answers:

5 0

When a substance gains heat, its molecules move more vigorously. When a substance loses heat, its molecular action decreases.

Conceptually and quantitatively, flow of energy during change of state effects motion of atoms and molecules in chemical and physical processes.

Phase change -

Now, the motion of a molecule can be increased by the following;

- Adding heat through a temperature difference

- The absorption of one of the five forms of energy

- Temperature increases according to specific heat of the substance

Nesterboy [21]3 years ago

4 0

All phase changes either have an increase or decrease of Intermolecular forces involved.

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Which of the gas laws takes into account the forces of attraction between molecules based on the characteristics of the gas? *

Alika [10]

The answer is Boyle’s Law

5 0

3 years ago

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

Mademuasel [1]

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of propane at $25.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

3 0

3 years ago

What is the difference between Nuclear Fission and Nuclear Fusion? Also any examples of them

stealth61 [152]

Answer:

Hiya there!

Explanation:

Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy.

For example, uranium can fission to yield strontium and krypton. Fusion joins atomic nuclei together. The element formed has more neutrons or more protons than that of the starting material. For example, hydrogen and hydrogen can fuse to form helium.

Hope this helped! :D

Credit sourced from "nuclear.duke-energy.com, thoughtco.com"

6 0

4 years ago

Which statement best describes the liquid state of matter?

KonstantinChe [14]

Answer:D. It has indefinite shape but definite volume.

Explanation: The property and characteristics of liquids is it does not have a definite shape since it copies and conform on the shape of its container but it has a definite of fixed volume.

6 0

3 years ago

Read 2 more answers

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago