What is meant by "the energy of an electron is quantized"?
2 answers:
Answer:
a and B are correct
Explanation:
The energy of an electron associated with one shell around the nucleus is quantized. And is given by the formula
.
where n= shell number 1,2,3...
It is true that energy of an electron cab be measured and Each electron around an atom has a discrete measure of energy associated with it. Moreover, quantity of electron energy remains constant in a shell and changes only when the electron changes its shell.
hence C is incorrect.
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Use the Clausius-Clapeyron equation...
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>![ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28P_1%2FP_2%29%20%3D%20%5Cfrac%7B-%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%20%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
<span>
</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>
</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>