After ionization, sodium gains a net positive charge cuz sodium loses its 1 valence electron to gain the nearest stable octet which is neon{Ne}. Hope it helps
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
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Answer:
3.711 L
Explanation:
The formula you need to use is the following:
3.4L / 298 K = V2 / 273 K
V2 = 3.711 L
We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
