<span><em><u>Climatology </u>is a <u>subspecialty </u>of a </em><u><em>Climate</em></u><em> and for how the <u>climate</u> changes. This is averaged out from over a set of a period of time.
<u>I hope this helps! ;D</u></em></span>
Answer:

Explanation:
The Rydberg equation gives the wavelength λ for the transitions:

where
R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

Data:

λ = 657 nm
Calculation:

Answer:
There is a lot of empty space between them
Explanation:
We know that gas molecules are loosely packed,
Therefore there is a lot of intermolecular space...(Which basically means that there is a lot of space between the molecules or particles in a gas)
Happy to help.
Pls mark as Brainliest.
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46