All categories

3 years ago

How are photons emitted by atoms?

Chemistry

2 answers:

Paul [167]3 years ago

6 0

Explanation:

<em>When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.</em>

Mrac [35]3 years ago

6 0

The photon is emitted with the electron moving from higher energy level to a lower energy level. It decreases energy and the atom emits

You might be interested in

What type of reaction provide energy within a red giant star?<br><br> and how come?

Dafna1 [17]

"A red giant is a star that has exhausted the primary supply of hydrogen fuel at its core. An average-sized star like our Sun will spend the final 10 percent of its life as a red giant. In this phase, a star's surface temperature drops to between 3,140 and 6,741°F (1,727 and 3,727°C) and its diameter expands to 10 to 1,000 times that of the Sun. The star takes on a reddish color, which is what gives it its name."

3 0

3 years ago

Balance the following equation. __AI+ __CuO __AI203+ _Cu

Nesterboy [21]

2 Al+ 3 CuO-> 1 Al2O3+ 3Cu

7 0

2 years ago

What parts are present in every chemical equation?

77julia77 [94]

I believe it’s

The reactants

The products

And the Arrow indicating the direction of the chemical reaction

3 0

3 years ago

What type of pollution in cities composed of car exhaust and industrial pollutes

Troyanec [42]

Air pollution is the answer

4 0

3 years ago

Read 2 more answers

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago