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Allushta [10]
3 years ago
15

What the concept of electronic transition​

Chemistry
1 answer:
joja [24]3 years ago
4 0

Answer: Electronic transition moments are defined as the probability for a given excitation energy transition to take place. It should be evident that the transition moment depends upon the spin-orbit coupling of the electrons in both the ground and excited states.

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What percentage of oxygen is attributed fossil fuel combustion
Oxana [17]

Answer:

C. 4%

Explanation:

The answer is 4%

6 0
2 years ago
The most condensed state of matter is <br> A. solid<br> B. liquid <br> C. gas <br> D. plasma
algol [13]
The most condensed state of matter is A. Solid Matter
6 0
3 years ago
Read 2 more answers
How many molecules of sulfur trioxide are in 78.0 grams?
zalisa [80]

Answer:

b) 5.87 E23 molecules

Explanation:

∴ mm SO3 = 80.066 g/mol

⇒ molecules SO3 = (78.0 g)(mol/80.066 g)(6.022 E23 molec/mol)

⇒ molec SO3 = 5.866 E23 molecules SO3

5 0
3 years ago
Please help, with step by step work
natali 33 [55]

\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

Henceforth –

\bf 3.60×10^{23} formula units \bf Cu_2O–

\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

5 0
2 years ago
if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti
ehidna [41]
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
7 0
4 years ago
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