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4 years ago

What are two important quantum-mechanical concepts associated with the Bohr model of the atom?

Chemistry

1 answer:

malfutka [58]4 years ago

4 0

Explanation:

The important quantum-mechanical concepts associated with the Bohr model of atom are :

1. Electrons are nothing but particles that revolve around the nucleus in discrete orbitals.

2. Energy is associated with each orbital is quantised. Meaning electron in each shell will have energy in multiple of a fixed quanta.

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What makes the soap feel soapy?

lapo4ka [179]

Answer:

because soap is a base.

Explanation:

bases are naturally bitter and slippery in nature.

6 0

2 years ago

Read 2 more answers

a sample of natural gas contains 8.24 moles of methane, 0.421 moles of ethane, and 0.116 moles of propane. if the total pressure

Varvara68 [4.7K]

The partial pressure (Px) of a gas in a gas mixture is equal to its mole fraction (Xi) multiplied by the total pressure (P) of the gas mixture. That means that we have to calculate the mole fraction of each gas, then calculate its partial pressure. The mole fraction of a gas is its number of moles (n) divided by the total number of moles.

$$ Mole fraction of methane: \\$\chi_{\text {methane }}=\frac{\mathrm{n}_{\text {methane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.24 \mathrm{~mol}+0.421 \mathrm{~mol}+0.116 \mathrm{~mol}} \chi_{\text {methane }}=\frac{8.24 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {methane }}=0.938$

$$Partial Pressure of methane:\\$\mathrm{P}_{\text {methane }}=\chi_{\text {methane }} \times \mathrm{PP}_{\text {methane }}=0.938 \times 1.37 \mathrm{~atm} \mathbf{P}_{\text {methane }}=\mathbf{1 . 2 8} \mathbf{~ a t m}$

$$Mole fraction of ethane: \\$\chi_{\text {ethane }}=\frac{\mathrm{n}_{\text {ethane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {ethane }}=\frac{0.421 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {ethane }}=0.0479$

$$Partial pressure of ethane:\\$\mathrm{P}_{\text {ethane }}=\chi_{\text {ethane }} \times \mathrm{PP}_{\text {ethane }}=0.0479 \times 1.37 \mathrm{~atm} \mathrm{P}_{\text {ethane }}=\mathbf{0 . 0 6 5 6} \mathbf{~ a t m}$

$$Mole fraction of propane:\\$\chi_{\text {propane }}=\frac{\mathrm{n}_{\text {propane }}}{\mathrm{n}_{\text {total }}} \chi_{\text {propane }}=\frac{0.116 \mathrm{~mol}}{8.78 \mathrm{~mol}} \chi_{\text {propane }}=0.0132$

<h3>What is Dalton’s Law?</h3>

Dalton's law of partial pressures is a gas law that states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture. The mole fraction of a given gas in a gas mixture is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gas mixture. This mole fraction can also be used to calculate the total number of moles of constituent gas if the total number of moles of the mixture is known. In addition, the mole fraction can also be used to calculate the volume of a certain gas in a mixtur.

To learn more about Dalton’s Law, visit:

brainly.com/question/14119417

#SPJ4

6 0

1 year ago

Consider the reaction cacn2 + 3 h2o → caco3 + 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced? 1. 2.13 mol 2. 63

kodGreya [7K]

Equation is as follow,

<span> CaCN</span>₂<span> + 3 H</span>₂<span>O → CaCO</span>₂<span> + 2 NH</span>₃

According to this equation,
When,
100 g CaCO₃ (1 mole) is produced when = 34 g NH₃ (2 moles) is produced
So,
187 g CaCO₃ will be produced then = X g of NH₃ will produce

Solving for X,
X = (187 g × 34 g) ÷ 100 g

X = 63.58 g of NH₃ will be produced

Result:
Option-2 is correct answer.

5 0

4 years ago

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl

Orlov [11]

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial molarity = 1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ = 0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

7 0

3 years ago

Assuming an efficiency of 41.50%, calculate the actual yield of magnesium nitrate formed from 136.7 g of magnesium and excess co

Sergio [31]

The actual yield is 489.8g of magnesium nitrate

5 0

3 years ago