Answer:
ii) the energy of the electron on the outer shell
iv) the overall size of an orbital
Explanation:
There are four quantum numbers to define the position and energy level of an electron in an atom
a) Principal : The principal quantum number (n) is to know the energy of an electron in an atom and its possible distance from the nucleus.
b) Azimuthal: It refers to the shape of the subshell or orbital of the electron and thus the angular distribution.
c) Magnetic: It refers to the number of orbits and their orientation in the subshell.
d) spin: It refers to the spin of the electron.
Answer:
![\Delta _{fus}H=3255.3J/mol](https://tex.z-dn.net/?f=%5CDelta%20_%7Bfus%7DH%3D3255.3J%2Fmol)
![\Delta _{fus}S=7.62\frac{J}{mol*K}](https://tex.z-dn.net/?f=%5CDelta%20_%7Bfus%7DS%3D7.62%5Cfrac%7BJ%7D%7Bmol%2AK%7D)
Explanation:
Hello,
Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:
![\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7BdT%7D%3D%5Cfrac%7B%5CDelta%20_%7Bfus%7DH%7D%7BT%5CDelta%20_%7Bfus%7DV%7D)
As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:
![p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )](https://tex.z-dn.net/?f=p_2-p_1%3D%5Cfrac%7B%5CDelta%20_%7Bfus%7DH%7D%7B%5CDelta%20_%7Bfus%7DV%7Dln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29)
Solving for the enthalpy of fusion we obtain:
![\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol](https://tex.z-dn.net/?f=%5CDelta%20_%7Bfus%7DH%3D%5Cfrac%7B%28p_2-p_1%29%28V_2-V1%29%7D%7Bln%28%5Cfrac%7BT_2%7D%7BT_1%7D%29%7D%20%3D%5Cfrac%7B%2811.84atm-1.00%20atm%29%28156.6cm%5E3%2Fmol-142.0cm%5E3%2Fmol%29%7D%7Bln%28%5Cfrac%7B429.26K%7D%7B427.15K%7D%20%29%7D%20%5C%5C%5C%5C%5CDelta%20_%7Bfus%7DH%3D32127.3atm%2Acm%5E3%2Fmol%2A%5Cfrac%7B101325Pa%7D%7B1atm%7D%2A%28%5Cfrac%7B1m%7D%7B100cm%7D%20%29%5E3%5C%5C%5CDelta%20_%7Bfus%7DH%3D3255.3J%2Fmol)
Finally the entropy of fusion is given by:
![\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}](https://tex.z-dn.net/?f=%5CDelta%20_%7Bfus%7DS%3D%5Cfrac%7B%5CDelta%20_%7Bfus%7DH%7D%7BT_1%7D%20%3D%5Cfrac%7B3255.3J%2Fmol%7D%7B427.15K%7D%5C%5C%20%5C%5C%5CDelta%20_%7Bfus%7DS%3D7.62%5Cfrac%7BJ%7D%7Bmol%2AK%7D)
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Answer:
Ratio of [CO₃²⁻] / [H₂CO₃] is 1.0x10⁻⁸
Explanation:
Based on the equilibriums:
H₂CO₃ ⇄ H⁺ + HCO₃⁻ Ka1 = 4.3x10⁻⁷
HCO₃⁻ ⇄ H⁺ + CO₃²⁻ Ka2 = 4.8x10⁻¹¹
The sum of both equilibriums is:
H₂CO₃ ⇄ 2 H⁺ + CO₃²⁻ K' = 4.3x10⁻⁷*4.8x10⁻¹¹ = 2.064x10⁻¹⁷
Where K' is defined as:
K' = 2.064x10⁻¹⁷ = [H⁺]² [CO₃²⁻] / [H₂CO₃]
If concentration of H⁺ is 4.5x10⁻⁵M:
K' = 2.064x10⁻¹⁷ = [4.5x10⁻⁵]² [CO₃²⁻] / [H₂CO₃]
1.0x10⁻⁸ = [CO₃²⁻] / [H₂CO₃]
Thus, <em>ratio of [CO₃²⁻] / [H₂CO₃] is 1.0x10⁻⁸</em>
Answer: D) Electrons
Explanation:
Daltons atomic theory was electrons
Answer:
D. All of the above
Explanation:
We can place a piece of glass tubing into rubber stopper after the tubing has been fire polished and cooled by
Lubricating the tubing with water or glycerin.
Using a towel or cotton gloves for protection.
or by
Twisting the tubing and stopper carefully.
Hence we can say that all the above methods are equally useful for the above procedure.