35.45
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
There are 3 moles of nitrogen in 3 moles of ammonium nitrate. Ammonium nitrate contains 2 nitrogen atoms in its chemical formula NH4NO3. Each mole of ammonium nitrate contains 2 moles of nitrogen atoms.
The balanced equation for the reaction is N2 + 3H2 --> 2NH3 Thus, the mole ratio of nitrogen to ammonia in the balanced equation is 1:2.
In dry air, nitrogen constitutes about 78% of the atmosphere by volume. Therefore, at 1 atmosphere pressure, the partial pressure of nitrogen would be 0.78 atm. This is calculated by multiplying the total pressure by the mole fraction of nitrogen in air.
To find the mass of nitrogen needed to make 34 g of ammonia, we first need to calculate the molar mass of ammonia (NH3), which is 17 g/mol. From this, we can see that 1 mole of ammonia contains 1 mole of nitrogen. Therefore, the mass of nitrogen needed would also be 34 g.
The confusion might arise because the term "mole" is a unit of measurement for the amount of substance, while "nitrogen" refers to a specific element. So saying "1 mole of nitrogen" could be ambiguous without specifying whether it refers to 1 mole of nitrogen atoms or 1 mole of nitrogen molecules (N2).
Number of nitrogen atoms in 1 mole nitrogen dioxide? Nitrogen dioxide has 1 N atom and 2 O atoms. One mole of nitrogen dioxide has 1 mole of N atoms
At standard temperature and pressure (STP) of 0°C and 1 atmosphere, 1 mole of any ideal gas occupies 22.4 liters of volume. This applies to nitrogen gas as well.
To find the mole fraction of oxygen, first convert the percentages to fractions: 37% oxygen is 0.37 and 63% nitrogen is 0.63. Since the total mole fraction in a mixture is 1, the mole fraction of oxygen would be 0.37/(0.37 + 0.63) = 0.37/1 = 0.37. Therefore, the mole fraction of oxygen in the gas mixture is 0.37.
There are 1 mole of nitrogen gas molecules contain 2 nitrogen atoms. Therefore, 0.25 mole of nitrogen gas would contain 0.25 * 2 = 0.5 moles of nitrogen atoms.
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
10 grams nitrogen (1 mole N/14.01 grams)(6.022 X 1023/1 mole N) = 4.3 X 1023 atoms of nitrogen ======================
To determine the volume of nitrogen needed to react with hydrogen, we need to know the balanced chemical equation. Once we have the balanced equation, we can use the stoichiometry of the reaction to calculate the volume of nitrogen. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L.
There are 3 moles of nitrogen in 3 moles of ammonium nitrate. Ammonium nitrate contains 2 nitrogen atoms in its chemical formula NH4NO3. Each mole of ammonium nitrate contains 2 moles of nitrogen atoms.
The balanced equation for the reaction is N2 + 3H2 --> 2NH3 Thus, the mole ratio of nitrogen to ammonia in the balanced equation is 1:2.
In dry air, nitrogen constitutes about 78% of the atmosphere by volume. Therefore, at 1 atmosphere pressure, the partial pressure of nitrogen would be 0.78 atm. This is calculated by multiplying the total pressure by the mole fraction of nitrogen in air.
There are approximately 3.01 moles of urea in 25 g of CONH2, so there are 3.01 moles of nitrogen atoms. Therefore, there are 3.01 moles * 2 nitrogen atoms/molecule = 6.02 moles of nitrogen atoms in 25 g of CONH2, or urea.