Explanation:
The given data is as follows.
T = 298 K, = -5645 kJ/mol
= -5798 kJ/mol
Relation between and are as follows.
=
-5798 kJ/mol = -5645 kJ/mol -
-153 kJ/mol = -
= 0.513 kJ/mol K
Now, temperature is = (37 + 273) K = 310 K
Since, =
=
= (-5645 kJ/mol - 159.03 kJ/mol)
= -5804.03 kJ/mol
As, change in Gibb's free energy = maximum non-expansion work
= -5804.03 kJ/mol - (-5798 kJ/mol)
= -6.03 kJ/mol
Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.
Answer:
*Inserting real answer*
Explanation:
I think CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4 is the skeleton equation.
If I am wrong I am so sorry!!!
Have a nice day! :)
Also sorry for all the fake answers you have been getting that happened to me too TnT
Answer:
the packing efficiency is 52.36%
Explanation:
Given the data in the question;
simple cubic unit cell that contains one atom with a metallic radius of 175 pm;
we know that;
Edge length of Simple cubic (a) is related to radius of atom (r) as follows;
a = 2r
since radius r = 175 pm
we substitute
a = 2 × 175 pm
a = 350 pm
Now we get the volume unit;
Volume of unit cell = a³ = ( 350 pm ) = 42875000 pm³
Next we get Volume of sphere;
Volume of Sphere = πr³
Volume occupied by 1 atom = × π × ( 175 pm )³
= × π × 5359375 pm³
= 22449297.5 pm³
Now, the packing efficiency = ( Volume occupied by 1 atom / Volume of unit cell ) × 100
we substitute;
packing efficiency = ( 22449297.5 pm³ / 42875000 pm³ ) × 100
= 0.523598 × 100
= 52.36%
Therefore, the packing efficiency is 52.36%
The relation between the volume and the temperature of the gas is given by Charles's law. The final temperature of the gas at 0.75 liters is -193.8°C.
<h3>What is Charles's law?</h3>
Charles's law was derived from the ideal gas equation and is used to state the relationship between the temperature and the volume of the gas. With a decrease in volume the temperature decreases.
If the pressure is kept constant then with an increase in temperature the volume of the gas expands. The law is given as,
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
Initial volume (V₁) = 2.80 L
Initial temperature (T₁) = 23 °C = 296.15 K
Final volume (V₂) = 0.75 L
Final temperature = T₂
Substituting the values above as:
T₂ = (V₂ × T₁) ÷ V₁
= 0.75 × 296.15 ÷ 2.80
= 79.325 K
Kelvin is converted as, 79.325K − 273.15 = -193.8°C
Therefore, the final temperature is -193.8°C.
Learn more about Charle's law, here:
brainly.com/question/16927784
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Explanation:
The given data is as follows.
= 2.34 kPa = 2.34 \times 1000 Pa = 2340 Pa = 0.0231 atm
= (20 + 273) K = 293 K
= 2537.4 kJ/kg = 2537400 J/kg
= ?, = (40 + 273) K = 313 K
According to Clausius-Clapeyron equation,
=
=
= atm
or, = 18846.45 kPa
Thus, we can conclude that the vapor pressure at is 18846.45 kPa.