Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .139 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.139-x ≈ 0.139
4.5x10^-4 = x^2/0.139
Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.007M/0.139M = .0503 or
≈5.03% dissociation.
<span>So to make it clear let's break the equation down species by species and assess the number of each species on bothe sides of the equation:
2C</span>₈H₈ + 25O₂ → 8CO₂ + 18H₂<span>O
LHS: C - 16 RHS: C - 8
H - 16 H - 36
O - 50 O - 34
Thus based on that it is evident that the equation is not quite balanced. This therefore means a "</span><span>No, because the number of carbon, hydrogen & oxygen atoms on both sides of the equation are not equal."
</span>The actual balance equation would be C₈H₈ + 10O₂ → 8CO₂ + 4H₂O
Answer:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
Explanation:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:
<u>Where:</u>
<em>μ (l): is the chemical potential of 2-propanol in solution </em>
<em>μ° (l): is the chemical potential of pure 2-propanol </em>
<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>
<em>T: is the temperature = 82.3 °C = 355.3 K </em>
<em>x: is the mole fraction of 2-propanol = 0.41 </em>
Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
I hope it helps you!
Answer:
12,375 cm³
Explanation:
v=l×w×h
Since there are 3 books the equation would then be
v=3(l×w×h)
v=3(25×15×11)
v=3(4,125)
v=12,375 cm³
