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pogonyaev
3 years ago
13

Microwave ovens emit microwave energy with a wavelength of 12.7 cm. what is the energy of exactly one photon of this microwave r

adiation?
Chemistry
1 answer:
Maksim231197 [3]3 years ago
4 0
The energy of a photon is equal to the  product of planck's constant,h and speed of light,c over the wavelength,y. Hence, e=h*\frac{c}{y}. h is equal to 6.6261*10^-34 Js, c is equal to 3*10^8 m/s and y= 12.7 cm or 0.127 m. Substituting, energy is equal to 1.5652*10^-24 Joules.
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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
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<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

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m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

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