Answer:
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Explanation:
Entropy is the measure of the degree of disorderness.
In solids, the entropy is very less compared to liquids and gases.
The entropy order is:
solids<liquids<gases
Among the given substances, water in liquid form has a strong intermolecular H-bond.
So, it has also less entropy.
Next acetic acid.
Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.
HCN has slight H-bonding in IT.
Hence, the entropy order is:
Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
Answer: the scientist should ensure that the volume is measured in dm3(or L), the pressure in atm, the temperature in Kelvin and use the gas constant 0.082atm.dm3.K^-1.mol^-1
Lets assume the gas is acting Ideally, then According to Ideal Gas Equation the density is given as,
d = P M / R T ------- (1)
Where;
P = Pressure = 1.03 atm
M = Molar Mass = 146.06 g/mol
R = Gas Constant = 0.08206 atm.L.mol⁻¹.K⁻¹
T = Temperature = 297 K
Putting Values in eq. 1,
d = (1.03 atm × 146.06 g/mol) ÷ (0.08206 atm.L.mol⁻¹.K⁻¹ × 297 K)
density = 6.17 g/L
