First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts
Answer: nn
Explanation:
The nanometre (international spelling as used by the International Bureau of Weights and Measures; SI symbol: nm) or nanometer (US spelling) is a unit of length in the metric system, equal to one billionth (short scale) of a metre (0.000000001 m).
Answer:
40.94 g
Explanation:
Given data:
Mass of NO₂ = ?
Volume = 20.0 L
Pressure = 110.0 Pka
Temperature = 25°C
Solution:
Pressure = 110.0 KPa (110/101 = 1.1 atm)
Temperature = 25°C (25+273 = 298.15 K)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.1 atm × 20.0 L / 0.0821 atm.L/ mol.K ×298.15 K
n = 22 / 24.5 /mol
n= 0.89 mol
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.89 mol × 46 g/mol
Mass = 40.94 g
Answer:
Nitrogen
Oxygen
Argon
Carbon Dioxide
Methane
Ozone
Explanation:
N₂ accounts for 78% of the atmosphere.
O₂ accounts for 21% of the atmosphere.
Ar accounts for 0.9% of the atmosphere.
CO₂, CH₄, and O₃ only take up 0.1% of the atmosphere.
