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Answer:
V = 22.41 L
Explanation:
Given data:
Mass of nitrogen = 14.0 g
Volume of gas at STP = ?
Gas constant = 0.0821 atm.L/mol.K
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles= 14 g/ 14 g/mol
Number of moles = 1 mol
Volume of gas:
PV = nRT
1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K
V = 22.41 atm.L / 1 atm
V = 22.41 L
Answer:
Recycling and reuse of materials
Explanation:
One of the greatest problems facing the human population is the problem of solid waste disposal. The menace of solid waste disposal has led to the idea of landfills. Land fills are depressions on the earth surface prepared for the purpose of solid waste disposal.
The most important approach towards solid waste disposal is the idea of recycling of materials. A material can be collected after use and processed into the same material or serve as a precursor in another manufacturing process. This means that no waste is generated as the materials which are supposed to be disposed of as solid waste are processed into other useful materials. This will reduce the volume of solid wastes generated that may need to be disposed in a landfill.
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
