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Mrac [35]
3 years ago
14

Calculate the pOH of this solution. pH = 1.90 pOH =

Chemistry
2 answers:
ladessa [460]3 years ago
8 0

Answer:

<h2>12.1 </h2>

Explanation:

Since the pH has been given, the pOH can be found by using the formula

pH + pOH = 14

pOH = 14 - pH

From the question we have

pOH = 14 - 1.9

We have the final answer as

<h3>12.1</h3>

Hope this helps you

Leni [432]3 years ago
7 0

Answer:

12.10

Explanation:

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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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