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2 years ago

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Which characteristic of water allows its molecules to form hydrogen bonds? The water molecule is polar. Water has a low specific

heat. Ionic bonds hold the water molecule together. Water expands as it freezes.

2 answers:

Fittoniya [83]2 years ago

5 0

Definitely the first: the water molecule is polar. :)

inysia [295]2 years ago

4 0

The water molecule is polar!

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A semi-infinite solid is

Anarel [89]

Answer:

D. infinitely extended in all directions

Explanation:

A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.

The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.

It is also expands as heat is applied because its thickness is negligible.

This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected

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Write the following as powers of ten with one figure before the decimal point:

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2 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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3 years ago

You have been asked to recommend a alternative source of energy that your town could use to generate electricity. Describe the r

snow_tiger [21]

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