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3 years ago

Increasing the temperature increases the vaporization rate of a liquid because the excess energy is used to break covalent bonds

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

False

Explanation:

False. The molecules of liquid are hold in the liquid state due to intermolecular forces or Van de Waals forces , without affecting the molecule itself and its atomic bonds (covalent bonds). When the temperature increases the kinetic energy of the molecules is higher , therefore they have more possibilities to escape from the attractive intermolecular forces and go to the gas state.

Note however that this is caused because the intermolecular forces are really weak compared to covalent bonds, therefore is easier to break the first one first and go to the gas state before any covalent bond breaks ( if it happens).

A temperature increase can increase vaporisation rate if any reaction is triggered that decomposes the liquid into more volatile compounds , but nevertheless, this effect is generally insignificant compared with the effect that temperature has in vaporisation due to Van der Waals forces.

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shepuryov [24]

Answer:

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Explanation:

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2 years ago

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

slega [8]

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$

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3 years ago

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Likurg_2 [28]

Answer:

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Explanation:

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2 years ago

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Kruka [31]

I’m pretty sure it’s B

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2 years ago

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ArbitrLikvidat [17]

Answer:

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2 years ago