Answer:
Orbital Notation is more specific on where exactly the electron is placed.
Explanation:
When writing an electron configuration for an atom, rather than writing out the occupation of each and every orbital specifically, you instead lump all the core electrons together and designate it with a symbol of the corresponding noble gas on the Periodic Table.
the arrangement of electrons in the orbitals of an atom or molecule
While Orbital Notation is a visual transformation of the electron configuration. It shows you where each specific electron is placed and what its "spin" is.
Glad I could help!
Answer:
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Explanation:
The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.
It works with molar concentrations on the equilibrium and it does not consider the solids compounds
Kc also can be modified by the time of the reaction.
This reaction is:
CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Answer:
Density of concentrated H2SO4 = 1.99g/cm^3 = 1991.79Kg/m^3
Explanation:
mass of empty flask = 78.23g mass of flask filled when with water = 593.63g.
mass of flask filled when with concentrateds sulfuric acid, H2SO4 = 1026.57g
Mass of water = (mass of flask filled when with water) -
(mass of empty flask) = 593.63g - 78.23g = 515.4g
Volume of flask = volume of water = volume of concentrateds sulfuric acid, H2SO4 =
(mass of water)/ density of water) = 515.4g/1.00g/cm^3 = 515.4cm^3
The density of concentrated sulfuric acid is given by
Density of concentrated H2SO4 = (mass of H2SO4) ÷ (volume of H2SO4) = 1026.57g/515.4cm^3 = 1.99g/cm^3 = 1991.79Kg/m^3
<u>The Concept:</u>
We are given the density of a sample of the metal = 11.4 grams / cm³
and we need to find the volume occupied by a sample of 30.5 grams
For this solution, we will use dimensional analysis
from the given information, we can also say that the density of the metal is:
1 cm³ / 11.4 grams
If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal
<u>Solving for the volume:</u>
X 30.5 grams = (30.5 / 11.4) cm³
Volume of 30.5 grams of the sample = 2.68 cm³
