A b and e is are the answers
Although you have not provided the circled electron, I can help you with a wide explanation.
1) Atomic number of manganese is 25. That means that it has 25 protons and 25 electrons.
2) Those 25 electrons are distributed (electron configuration) as per the quantum rules:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
3) The most reasonable is that you have been asked to give the possible quantum numbers for an electron in the 4s or 3d.
4) Those are 7 electrons and these are their possible sets of quantum numbers:
i) For the two electrons in 4s:
n is the main energy level so n = 4
l tells the kind of orbital, which is s, so l = 0
ml is also 0 (it can be from -l to + l, so given that l i s0, ml is 0)
ms: one is +/12 and the other is -1/2 (this is the spin number).
ii) For the 5 electrons in 3d
n = 3
l can be 0, 1, or 2
if l = 0, then ml = 0
if l = 1, then ml can be -1, 0 , or 1 (from - l to + l)
ms can be either +1/2 or - 1/2 (spin)
Answer:
6M
Explanation:
(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted doln
Molarity dilute soln = [(M x V)conc/V (dilute)] = 1.5L x 12M / 3.0L = 6M final dilute soln
Answer:
Explanation:
If l = 3, the electrons are in an f subshell.
The number of orbitals with a quantum number l is 2l + 1, so there
are 2×3 + 1 = 7 f orbitals.
Each orbital can hold two electrons, so the f subshell can hold 14 electrons.
Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
