Input interpretation
oxygen + nitrogen ⟶ N2O2
Balanced equation
Balance the chemical equation algebraically: + ⟶ N2O2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 N2O2 Set the number of atoms in the reactants equal to the number of atoms in the products for O and N: O: | 2 c_1 = 2 c_3 N: | 2 c_2 = 2 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | + ⟶ N2O2
Structures
+ ⟶ N2O2
Names
oxygen + nitrogen ⟶ N2O2
Chemical names and formulas
| oxygen | nitrogen | N2O2 formula | | | N2O2 Hill formula | O_2 | N_2 | N2O2 name | oxygen | nitrogen | IUPAC name | molecular oxygen | molecular nitrogen |
Substance properties
| oxygen | nitrogen | N2O2 molar mass | 31.998 g/mol | 28.014 g/mol | 60.012 g/mol phase | gas (at STP) | gas (at STP) | melting point | -218 °C | -210 °C | boiling point | -183 °C | -195.79 °C | density | 0.001429 g/cm^3 (at 0 °C) | 0.001251 g/cm^3 (at 0 °C) | solubility in water | | insoluble | surface tension | 0.01347 N/m | 0.0066 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | odorless | odorless |
Units
