It would have to increase pressure... but I don’t see that option here..?
Answer:
The probability density (ψ2)
Explanation:
Indicates the probability of finding the electron in a certain region of space when it is squared ψ2.
This means that define2 defines the distribution of electronic density around the nucleus in three-dimensional space; a high density represents a high probability of locating the electron and vice versa.
The atomic orbital, can be considered as the electron wave function of an atom.
APPLICATIONS:
1.- Specify the possible energy states that the electron of the hydrogen atom can occupy and identify the corresponding wave functions medio, by means of a set of quantum numbers, with which an understandable model of the hydrogen atom can be constructed.
2.- It does not work for atoms that have more than one electron, but the problem is solved using approximation methods for polyelectronic atoms.
Answer:
The hydrogen atom has just one electron, but many spectral lines. However it contains many shells and the movement of that electron from one shell to another causes the release of energy and also an emission of photons.
A spectral line are dark or bright lines formed within a specific frequency range which differ from other frequencies.Because of the difference of energy for the various shells, it produces different wavelengths and this is the reason for the many spectral line for hydrogen.
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
