Answer:
Taxonomy
Explanation: Taxonomy formally names and classifies organisms by thier structure, function, and relationships
The chemical equation needs to be balancedso that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side. Balancing chemical equations is a process of trial and error. hope this helps
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.
At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.
The mole fraction of nitrogen in the mixture is:
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
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Answer:
P₂ = 5.550213 Pa
Explanation:
Given data:
Heat of vaporization = 376.6 KJ/mol ( 376600 j/mol)
R = 8.3143 j mol⁻¹ K⁻¹
Temperature = T1 = 100 °C = 100 + 273 = 373 K
Temperature = T2 = 508.0 °C = 508.0+273 = 781 K
Pressure at 100°C= P1 = 1.6 × 10⁻³² Pa
Pressure at 508.0 °C = ?
Solution:
ln P₁/P₂ = ΔH /R ( 1/T₂ - 1/T₁)
ln (1.6 × 10⁻³² Pa) - ln (P₂) = 376600 j. mol⁻¹ / 8.3143 j mol⁻¹ K⁻¹ ( 1/T₂ - 1/T₁)
-73.212719 - ln (P₂) = 45295.45 (0.001280 - 0.002680)
-73.212719 - ln (P₂) = 45295.45 (-0.00414)
- ln (P₂) = -63.41363 + 73.212719
- ln (P₂) = 9.799089
P₂ = e⁻⁹°⁷⁹⁹⁰⁸⁹
P₂ = 5.550213 Pa