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MissTica
2 years ago
15

How many moles of hydrogen atoms are in 1 mole of sulfuric acid?

Chemistry
1 answer:
Andrews [41]2 years ago
7 0

Answer:

2 moles

Explanation:

In one mole of sulfuric acid

There is 2 moles of hydrogen

4 moles of oxygen

and 1 mole of sulfur.

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When studying atoms, scientists can ignore the
vekshin1

When studying atoms, scientists can ignore <u>the Gravitational</u> force between charged particles that make up  the atoms because it is many millions of times smaller than other forces in the atom.

Explanation:

Scientists can ignore the gravitational force because the gravitational force is considered to be negligible as compared to the other forces due to its smaller value.We all know that the gravitational force is directly proportional to the mass of an object which  result in  a small force value.When the value of this small force is compared to the value of the electrical force between protons and electrons in atoms the we can say that the electrical force is million times stronger than the gravitational force

Thus we can say that scientists can ignore <u>the Gravitational</u> force between charged particles that make up  the atoms because it is many millions of times smaller than other forces in the atom.

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3 years ago
How many grams of Hydrogen (H) would need to react with 143 grams of Oxygen (O) to make 235 grams of Water (H20)? 235
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45 grams

Explanation:

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2 years ago
Some common substances used in the laboratory are listed in the table. the chemical formulas of the substances are also listed b
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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
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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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