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4 years ago

When boiling water is the particles expanding?

1 answer:

5 0

When water is boiling the particles in the liquid are not expanding. This also applies for other liquids. Particles do not expand, it is only the volume they take up that expands. When water is boiling, the particles are rather escaping the liquid phase by undergoing a phase change which forms the particles into the gaseous phase.

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It’s d I took the test a few day ago

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3 years ago

PLEASE ANSWER ASAP What does a ball falling through the air have?

max2010maxim [7]

Answer:

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Explanation:

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3 years ago

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How many significant digits are in the number 5.40

alexdok [17]

There are 3 significant digits in 5.40.

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3 years ago

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The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is blank liters

Aneli [31]

Answer: 18.97 L

Explanation:

This can be solved by the Ideal Gas equation:

$P.V=n.R.T$

Where:

$P=2 atm$ is the pressure of the gas

$V$ is the volume of the gas

$n=1.5 mole$ the number of moles of gas

$R=0.0821\frac{L.atm}{mol.K}$ is the gas constant

$T=35\°C+273.15=308.15 K$ is the absolute temperature of the gas in Kelvin

Finding $V$ :

$V=\frac{nRT}{P}$

$V=\frac{(1.5 mole)(0.0821\frac{L.atm}{mol.K})(308.15 K)}{2 atm}$

$V=18.97 L$

Therefore:

The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is 18.97 liters

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4 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

olga_2 [115]

Answer: The $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

Explanation:

For the given half reactions:

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V$

Net reaction: $Fe(s)+Cu^{2+}\rightarrow Fe^{2+}+Cu(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34-(-0.44)=0.78V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.78 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}$

Hence, the $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

3 0

4 years ago