Answer:
n = 4, l = 2
Explanation:
The number 4 in 4d is the principal quantum number (n).
The letter d in 4d tells us that we have a d orbital, as determined by the <em>secondary quantum number (l</em>).
The quantum number l tells us the shape of the orbital.
l = 0 s orbital
l = 1 p orbital
l = 2 d orbital
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Answer:
Check the explanation
Explanation:
When talking about our universe there are 5 d orbitals. The element of first transition series moves away from the universal principles of Hund's rule and Aufbav's principle. So in order to attain stability these elements tend to form half or full filled orbitals.
In our universe the ground state electronic configuration of sixth transition metal, Iron (Fe) : [Ar] 
and the electronic configuration of seventh transition metal, Cobalt (Co) : [Ar] 
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In universe L there are seven orbitals.
Ground state electronic configuration of sixth and seven transition element.
Sixth transition metal: [Ar] ![3d^{7} 4s^1 or [X] 3d^{7} 4s^1](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E1%20or%20%5BX%5D%203d%5E%7B7%7D%204s%5E1)
Seventh transition metal: [Ar] ![3d^{7} 4s^{2}or [X] 3d^{7} 4s^{2}](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E%7B2%7Dor%20%5BX%5D%203d%5E%7B7%7D%204s%5E%7B2%7D)
Answer:
It will decrease by 2 units.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
pH = pKa + log(base/acid)
Let's assume your acid has pKa = 5.
(a) If the base: acid ratio is 1:1,
pH(1) = 5 + log(1/1) = 5 + log(1) = 5 + 0 = 5
(b) If the base: acid ratio is 1:100,
pH(2) = 5 + log(1/100) = 5 + log(0.01) = 5 - 2 = 3
(c) Difference
ΔpH = pH(2) - pH(1) = 5 - 3 = -2
If you increase the acid:base ratio to 100:1, the pH will decrease by two units.
