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mezya [45]
3 years ago
13

Be sure to answer all parts. Select the anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base. mc2x mc2xd1 mc2

xans mc2xd2 Then draw the product that results after elimination of HBr. draw structure ...
Chemistry
1 answer:
sleet_krkn [62]3 years ago
6 0

The anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base is shown in the image attached as well as the structure of the product formed in the reaction.

In organic chemistry, an antiperiplanar conformation is one in which the groups point up and down at a dihedral angle of 180° away from one another. In the image attached, the antiperiplanar conformation of (CH3)2CHCH2Br is shown.

Recall that an E2 reaction is a synchronous elimination reaction where to atoms leave at the same time. The product of this reaction is also shown in the image attached.

Learn more: brainly.com/question/2510654

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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}

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Answer:

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