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WINSTONCH [101]
3 years ago
5

How much energy in kJ is associated with a radio wave of wavelength 1.2 X 10^2 m?

Chemistry
1 answer:
eduard3 years ago
6 0
It's very simple.

Energy = \frac{hc}{\lambda}

where, h= Planck's constant = 6.6 x 10^{-34} m^2 kg/s

c= speed of light = 3 x 10^{8} m/s
\lambda = wavelength


So, energy = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{120}

= 1.65 x 10^{-27} J
=1.65 x 10^{-30} KJ

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4 protons the number of proton has the same number of electron [which is the same as atomic number]
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What is the velosity of a 72.3 kg jogger with a kinetic energy of 1080.0 J?
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Answer: 5.47m/s

Explanation:

Mass = 72.3kg

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1 year ago
How many molecules are in 1 mole of H2O
Olegator [25]

Answer:

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8 0
2 years ago
Suppose a solution is described as concentrated. Which of the following statements can be concluded? Select the correct answer b
Novosadov [1.4K]

Answer:

  • last option: none of<u> the above.</u>

Explanation:

Describing a solution as<em> concentrated</em> tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.

One conclusion, of course, is that <u>the solute is soluble</u>: else the solution were not concentrated.

On the other hand, the terms saturated and <em>supersaturated</em> to define a solution are specific.

A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A <u>supersaturated solution has more solute dissolved than the saturated solution</u> at the same temperature; superstaturation is a very unstable condition.

From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is<u> none of the above</u>.

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3 years ago
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