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tester [92]
3 years ago
7

What is the energy of a photon with a 6 micrometer wavelength (1 m = 106 micrometers)?

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
6 0
Hi hi I hope you understand that I’m gonna was the day u
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In a problem, you are given two pressures and one temperature at constant volume and amount of gas. You are asked to find a seco
laiz [17]

I don't know who it was named for but the law is

P1/T1 = P2/T2

Make sure the pressure units are the same (atmospheres or kPa usually) and that the temperature is in Degrees Kelvin which is derived from Celsius degrees.

Try Charles' Law for the name.

3 0
3 years ago
Read 2 more answers
What is the name of the ionic compound RbCI
lorasvet [3.4K]
It would be Rubidium chloride.
6 0
3 years ago
Which condition causes a hurricane to rotate?
Montano1993 [528]

Answer:

its caused by the convection of air masses with differences in densities mainly due to their differences in temperatures.

-Hops

7 0
2 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
2 years ago
A 6.50x10^-5 m solution of potassium permanganate has a percent transmittance of 27.3% when measured in a 1.15 cm cell at a wave
mihalych1998 [28]
For this problem, we use the Beer Lambert's Law. Its usual equation is:

A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
C is the concentration of the sample solution

As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:

A = 2 - log(%T)
A = 2 - log(27.3)
A = 0.5638
7 0
2 years ago
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