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4 years ago

What happens to an electron when it emits a photon

Chemistry

1 answer:

Flura [38]4 years ago

7 0

Answer:

when electron emit the radiations it means it jumped to the lower energy level from higher energy level.

Explanation:

When electron jump into lower energy level from high energy level it loses the energy.

The process is called de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

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A solution of 400mg of optically active 2-butanol in 10 mL of water, placed in a 20 cm cell shows an optical rotation of 40o. Wh

labwork [276]

Answer:

Specififc rotation [∝] = 0.5° mL/g.dm

Explanation:

Given that:

mass = 400 mg

volume = 10 mL

For a solution,

The Concentration = mass/volume

Concentration = 400/10

Concentration = 40 g/mL

The path length l = 20 cm = 2 dm

Observed rotation [∝] = + 40°

Specififc rotation [∝] = ∝/l × c

where;

l = path length

c = concentration

Specififc rotation [∝] = (40 / 2 × 40)

Specififc rotation [∝] = 0.5° mL/g.dm

3 0

3 years ago

The volume of a gas filled balloon is 22L at 313K and 1.5 atm. What would the volume be at 273K and 1 atm?

avanturin [10]

From the calculations and the statement of the general gas equation, the volume of the gas becomes 28.8 L

The gas laws are used to show the relationship between the variables in problems that has to do with gases.

In this case, we have to apply the general gas law as follows;

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = * 22L * 1.5 atm * 273K/313K * 1 atm

V2 = 28.8 L

Learn more about gas laws:brainly.com/question/12667831

#SPJ2

7 0

3 years ago

Read 2 more answers

What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m

patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

7 0

3 years ago

Rango carefully mixed 15.6 g of sodium with chlorine. The reaction produced 39.7 g of sodium chloride. How many grams of chlorin

Evgen [1.6K]

Answer:

24.1g of chlorine & 15.6g of sodium

Explanation:

according to the law of conservation of mass

Na+Cl=NaCl only

5 0

4 years ago

A 1.40 L sample of O2 at 645 Torr and 25 °C, and a 0.751 L sample of N2 at 1.13 atm and 25 °C, are both transferred to the same

anyanavicka [17]

Answer:

P(O₂) = 0.595 atm

P(N₂) = 0.424 atm

Total Pressure = 1.019 atm

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to <u>calculate the moles of each one</u>:

O₂:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

N₂:

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can <u>calculate the partial pressure for each gas in the new container</u>, because the number of moles did not change:

O₂:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

N₂:

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to <u>calculate the total pressure</u>:

Pt = 0.595 atm+ 0.424 atm = 1.019 atm

6 0

3 years ago