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Step2247 [10]
3 years ago
6

Consider the reaction below. If you start with 4.00 moles of C3H8 (propane) and 4.00 moles of O2 , how many moles of propane wou

ld be consumed?
Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer:

0.800 mol

Explanation:

We have the amounts of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with moles of the compounds involved.  

Step 1. <em>Gather all the information</em> in one place.

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:  4.00    4.00

===============

Step 2. Identify the <em>limiting reactant </em>

Calculate the <em>moles of CO₂</em> we can obtain from each reactant.  

<em>From C₃H₈:</em>

The molar ratio of CO₂: C₃H₈ is 3:1

Moles of CO₂ = 4.00 × 3/1

Moles of CO₂ = 12.0 mol CO₂

<em>From O₂</em>:

The molar ratio of CO₂: O₂ is 3:5.

Moles of CO₂ = 4.00 × ⅗

Moles of CO₂ = 2.40 mol CO₂

O₂ is the limiting reactant because it gives the smaller amount of CO₂.

==============

Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.

The molar ratio of C₃H₈:O₂ is 1:5.

Moles of C₃H₈ = 4.00 × ⅕

Moles of C₃H₈ = 0.800 mol C₃H₈

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3. Sulfur
musickatia [10]

Explanation:

 We are to write the orbital notation and electronic configuration of sulfur and iron.

The orbital notation shows the filling of electrons into orbitals or sublevels.

Electron configuration  shows the distribution of electrons into shells;

        Number of electrons     Electron configuration     Orbital notation

S                  16                                   2 8 6                      1s² 2s² 2p⁶ 3s² 3p⁴

Fe                26                                2, 8, 14 2           1s² 2s² 2p⁶ 3s² 3p⁶3d⁶4s²        

3 0
3 years ago
an electron in the 3rd shell of an Aluminium atom moves to the first shell in a bombardment process. Calculate the frequency of
Liula [17]

Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

<h3>What is quantum mechanic?</h3>

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in <em>Joules </em>or <em>electroVolts </em>

  • E₁ = -2.18x10⁻¹⁸ J
  • E₁ = -13.6 eV

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

  • E₁ =  -2.18x10⁻¹⁸ J / 1² =<u>  -2.18x10⁻¹⁸</u><u> J</u>

        E₁ =  -13.6 eV / 1² =<u>  -13.6 </u><u>eV</u>

  • E₃ =  -2.18x10⁻¹⁸ J / 3² =  -2.18x10⁻¹⁸ J / 9 =<u> - 2.42x10⁻¹⁹ </u><u>J</u>

        E₃ =  -13.6 eV / 3² =  -13.6 eV / 9 = <u>- 1.51 </u><u>eV</u>

The change of energy can be calculated in two ways,

<u>Option 1</u>

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ =<u> 1.93x10⁻¹⁸</u><u>J</u>

ΔE = E₁ - E₃ = 13.6 - 1.51 = <u>12.09 </u><u>eV</u>

<u>Option 2</u>

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

  • ni = 3
  • nf = 1

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

<u>ΔE</u><u> = - 1.937x10⁻¹⁸ </u><u>J</u>

or

ΔE = -13.6 eV  (1/nf² - 1/ni²)

ΔE = -13.6 eV  (1/1² - 1/3²)

ΔE = -13.6 eV  (1 - 0.111)

ΔE = -13.6 eV  (0.888)

<u>ΔE</u><u> = -12.08</u><u> eV</u>

This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

<u />

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s =  <u>2.923x10¹⁵ Hz</u>.

<u>Answer</u>:

  • Frequency, ν = E/h = <u>2.923x10¹⁵ </u><u>Hz</u>.
  • Energy, E = <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

You can learn more about quantum mechanic at

brainly.com/question/11855107

brainly.com/question/23780112

brainly.com/question/11852353

5 0
1 year ago
If boron atoms have 5 electrons, how many electrons are in the outer electron shell of a boron atom?
ANEK [815]

Answer:

3

Explanation:

The first election shell can only hold 2 electrons, but the next one can hold up to 8

3 0
2 years ago
Consider this reaction:
nadya68 [22]
11.7 g hope this helps and have a great day
4 0
2 years ago
How many miles are there in 20g of KClO3 (k=39 Cl=35.5 O=16)​
SashulF [63]
<h3>Answer:</h3>

0.144 moles

<h3>Explanation:</h3>
  • The relationship between mass of a compound, number of moles and molar mass of the compound is given by;
  • Number of moles = Mass ÷ Molar mass
  • Molar mass is equivalent to the relative formula mass of the compound that is calculated the atomic masses of the elements making the compound.

In this case;

Our compound, KClO3 will have a molar mass of;

= 39 + 35.5 + 4(16)

= 138.5 g/mol

Mass of KClO3 is 20 g

Therefore;

Number of moles = 20 g ÷ 138.5 g/mol

                            = 0.144 moles

Thus, the number of moles in 20 g of KClO3 is 0.144 moles

4 0
3 years ago
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