Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer:6 atoms
Explanation:Each NaOH has one Na and one O and one H. Therefore, 2 NaOH has 6 atoms.
Answer:
C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.
Explanation:
The Hund's rule is used to place the electrons in the orbitals is it states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied;
2. All of the electrons in singly occupied orbitals have the same spin.
So, the electrons first seek to fill the orbitals with the same energy (degenerate orbitals) before paring with electrons in a half-filled orbital. Orbitals doubly occupied have greater energy, so the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, and for the second statement, they have the same spin.
The other alternatives are correct, but they're not observed by the Hund's rule.
