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egoroff_w [7]
3 years ago
8

If the mass of a radioactive substance is 8 grams and it has a half-life of 4 hours, how much mass remains after 8 hours?

Chemistry
2 answers:
lisov135 [29]3 years ago
6 0

Answer: 2 grams will be remaining.

Explanation: Half life is the time in which the amount of radioactive substance remains half.

For example if half life of a radioactive substance is 3 hours and we have 16 grams of it then in two hours, 8 grams will be remaining. In next two hours, 4 grams will be remaining. Similarly, in next two hours, 2 grams and further in next two hours 1 gram of the substance will be remaining.

For the given problem, the half life of the radioactive substance is 4 hours. It means the initial amount of this substance will remain half in 4 hours.

Originally we have 8 grams of the radioactive substance. In first 4 hours, 4 grams of the substance will be remaining. Now, in next 4 hours that is in total 8 hours, 2 grams of the substance will be remaining.

These problems could also be solved mathematically using the formula:

\frac{N}{N_0}=(0.5)^n

where, N_0 is the initial amount of the substance, N is remaining amount and n is the number of half lives.

number of half lives(n) = \frac{time}{half life}

From given information, time is 8 hours and half life is 4 hours.

So, n = \frac{8}{4}  = 2

Initial amount is given as 8 grams and it asks to calculate the remaining amount. Let's plug in the values in the equation:

\frac{N}{8}=(0.5)^2

\frac{N}{8} = 0.25

N = 8(0.25)

N = 2

So, from both ways 2 grams of the radioactive substance will be remaining after 8 hours.

Alborosie3 years ago
3 0
The sample will lose half of its mass after 4 hours. The half life.
82=4
The sample will lose half of the remaining four after another half life.
42=2      Hope this helps! :)
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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                          

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  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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