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melisa1 [442]
3 years ago
7

Atoms have been traditionally viewed as being composed of three different types of particles: protons, neutrons, and electrons.

However, research done by Louis de Broglie and expanded upon by Erwin Schrödinger changed how scientists viewed electrons. How has our increased understanding of the electron led to current atomic models?
Chemistry
1 answer:
sattari [20]3 years ago
7 0

Answer:

See explanation

Explanation:

Before the advent of the wave-particle duality theory proposed by Louis de Broglie, there was a sharp distinction between mater and waves.

However, Louis de Broglie introduced the idea that mater could display wave-like properties.  Erwin Schrödinger developed this idea into what is now known as the wave mechanical model of the atom.

In this model, electrons are regarded as waves. We can only determine the probability of finding the electron within certain high probability regions within the atom called orbitals.

This idea has been the longest surviving atomic model and has greatly increased our understanding of atoms.

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A gas mixture is composed of three different gases. The first gas in the mixture has a pressure of 0.35 atm. The second gas in t
riadik2000 [5.3K]

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1.20atm

Explanation:

Given parameters:

Partial pressure of gas 1 = 0.35atm

Partial pressure of gas 2  = 0.20atm

Partial pressure of gas 3  = 0.65atm

Unknown:

Total pressure of the gas mixture = ?

Solution:

To solve this problem, we need to recall and understand the Dalton's law of partial pressure.

Dalton's law of partial pressure states that "the total pressure of a mixture of gases is equal to the sum of the partial pressure of the constituent gases".

    Total pressure  =Pressure of gas(1 + 2 + 3)

The partial pressure is the pressure a gas would exert if it alone occupied the volume of the gas mixture.

     

Now we substitute;

      Total pressure  = (0.35 + 0.20 + 0.65)atm  = 1.20atm

8 0
3 years ago
Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer
Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is  K_c= 2.8*10^{-4}

Explanation:

      From  the question we are told that

              The chemical reaction equation is

      Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)}  -----> 2Fe_{(s)} + 3H_{2} O_{(g)}

The voume of the misture is  V_m = 5.4L  

  The molar mass of  Fe_{2} O_{3}_{(s)} is a constant with value of  M_{Fe_{2} O_{3}_{(s)} } = 160g/mol

    The molar mass of  H_{2}_{(g)}    is a constant with value of  H_2 = 2g/mol

   

    The molar mass of  H_{2}O    is a constant with value of  H_2O = 18g/mol

Generally the number of moles  is mathematically given as

                     No \ of \ moles \ = \frac{mass}{molar\  mass}

    For   Fe_{2} O_{3}_{(s)}

          No \ of\ moles = \frac{3.54}{160}

                                = 0.022125 \ mols

     For  H_{2}

               No \ of\ moles = \frac{3.63}{2}

                                = 1.815 \ mols

       For  H_{2}O

                         No \ of\ moles = \frac{2.13}{18}

                                              = 0.12 \ mols

Generally the concentration of a compound  is mathematicallyrepresented  as

       Concentration  = \frac{No \ of \ moles }{Volume }

      For   Fe_{2} O_{3}_{(s)}

                Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}

                                         = 4.10*10^{-3}M                          

       For  H_{2}

                  Concentration[H_2] = \frac{1.815}{5.4}

                                           = 0.336M

      For  H_{2}O

                Concentration [H_2O] = \frac{0.12}{5.4}

                                                  = 0.022M

  The equilibrium constant  is mathematically represented as

                K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}

  Considering H_2O  \ for \ product

            And      H_2  \ for  \ reactant

At  equilibrium the

                    K_c = \frac{0.022}{0.336}

                          K_c= 2.8*10^{-4}

3 0
3 years ago
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