The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Answer:
Fe(NO3)3 + 3 NaOH ===》Fe(OH)3 + 3 NaNO3
Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
True.......................................
