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kolezko [41]
3 years ago
6

Based on the research of Albert Einstein, what change would most likely result in stopping the emission of electrons from this m

etal?
an increase in the intensity of the light
a decrease in the intensity of the light
the use of light that has a higher frequency
the use of light that has a lower frequency
Chemistry
2 answers:
saul85 [17]3 years ago
7 0
The use of light that has a lower frequency
ANEK [815]3 years ago
5 0

Answer:  The use of light that has a lower frequency

Explanation:  In order to stop the emission of electrons from the metal, the light used should be of lower frequency so that it wont be able to strike off the metal atom as effectively as it could.

When we use the light of lower frequency, the kinetic energy of the light decreases and thus it wont be able to emit much of the electrons from the metal surface.

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Entalpy of vaporization of water is 41.1k/mol. if the vapor pressure of water at 373k is 101.3 kpa, what is the vapor pressure o
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Answer: The vapor pressure of water at 298 K is 3.565kPa.

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

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

where,

P_1 = initial pressure at 298 K = ?

P_2 = final pressure at 373 K = 101.3 kPa

\Delta H_{vap} = enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 298 K

T_2 = final temperature = 373 K

Now put all the given values in this formula, we get

\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}]

\frac{101.3}{P_1}=antilog(1.448)

P_1=3.565kPa

Therefore, the vapor pressure of water at 298 K is 3.565kPa.

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