What is necessary to produce visible light from chemical compounds?

Chemistry

1 answer:

erastova [34]3 years ago

5 0

Answer:

the molecule must contain either pi bonds or atoms with non-bonding orbitals.

Explanation:

A non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen.

You might be interested in

Label A is showing___ frequency waves, and label B is showing ___ frequency waves

igomit [66]

It's not really possible to tell longitudinal vs. transverse in this image as given. However, we can say that the waves labeled A are high-frequency (short wavelengths) while the waves labeled B are low-frequency (long wavelengths). So, this third answer choice would be correct here.

6 0

3 years ago

a 5L container contains 3 moles of helium and 4 moles of hydrogen at a pressure of 9 atms maintaining a constant T and additiona

Stells [14]

Answer:

7.71 atm

Explanation:

Given the following data:

$V = 5 L$

$n_{He} = 3 mol$

$n_{H_2} = 4 mol$

$p_1 = 9 atm$

$T = const$

According to the ideal gas law, we know that the product between pressure and volume of a gas is equal to the product between moles, the ideal gas law constant and the absolute temperature:

$pV = nRT$

Since the temperature and the ideal gas constant are constants, as well as the fixed container volume of 5 L, we may rearrange the equation as:

$\frac{p}{n}=\frac{RT}{V}=const$

This means for two conditions, we'd obtain:

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

Given:

$p_1 = 9 atm$

$n_1 = n_{initial total} = n_{He} + n_{H_2} = 3 mol + 4 mol = 7 mol$

$n_2 = n_{final total} = n_{He} + n_{H_2} = 3 mol + 4 mol + 2 mol = 9 mol$

Solve for the final pressure:

$p_2 = p_1\cdot \frac{n_2}{n_1}$

Now, according to the Dalton's law of partial pressures, the partial pressure is equal to the total pressure multiplied by the mole fraction of a component:

$p_{H_2}=\chi_{H_2}p_2$