If 4.20 mol of ethane, C2H6, undergo combustion according to the unbalanced equation given below, how many moles of oxygen are r
1 answer:
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Answer:
1 - Salts required that can be compared directly, using Ksp values, to estimate solubilities for copper (II) sulfide are - (Option b) ,
(Option c) and
(Option d)
2 - Salts that can be compared directly, using Ksp values, to estimate solubilities for zinc hydroxide are - (Option a)
Explanation:
values can be used to compare the solubility of salts that produce ions in a 1:1 ratio.
{copper (II)sulfide} dissociates with ions in ratio 1:1 .
Because the salts ,
and
have a 1:1 ion ratio, the solubilities of options b, c, and d can be compared to salt 1.
Values can be used to compare the solubility of salts that produce ions in a 1:2 or 2:1 ratio.
2. dissociates into ions in ratio 1:2 .
Because the ions in the salt are in a 1:2 ratio, the solubility of salt 2 can be compared to that of salt 'a'.
Hence , Salt 1 is matched with options b , c , d.
Salt 2 is matched with option a.
Answer:
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:
With further simplification, we have:
Then, we have:
Therefore,
Using the expansion:
Therefore,
Thus:
equation (1)
Using the virial equation of state:
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2