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BARSIC [14]
3 years ago
15

Uranium has atomic number 92. Its most common isotope is 238U, but the form used in nuclear bombs and nuclear power plants is 23

5U. How many neutrons are in each of these two isotopes of uranium
Chemistry
1 answer:
Yakvenalex [24]3 years ago
3 0

Explanation:

Atomic Number = Number of protons

Mass Number = Number of protons + Number of neutrons

Isotopes are simply atoms of an element with the same number of protons and different number of neutrons.

First Isotope -- 238U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons =  238 - 92 = 146

Second Isotope -- 235U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons =  235 - 92 = 143

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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank
Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is K_c  = 14.39

Explanation:

The chemical equation for this decomposition of ammonia is

                2 NH_3  ↔   N_2 + 3 H_2

The initial concentration of ammonia is mathematically represented a

          [NH_3] =  \frac{n_1}{V_1}  = \frac{29}{75}

          [NH_3] = 0.387  \  M

The initial concentration of nitrogen gas  is mathematically represented a

         [N_2] =  \frac{n_2}{V_2}

         [N_2] =  0.173  \  M

So  looking at the equation

   Initially (Before reaction)

      NH_3 = 0.387 \ M

      N_2  =  0 \  M

      H_2 =  0 \ M

During reaction(this is gotten from the reaction equation )

        NH_3 = -2 x(this implies that it losses two moles of concentration )

         N_2 = + x  (this implies that it gains 1 moles)

         H_2  =  +3 x(this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

        NH_3 = 0.387 -2x

       N_2 =  x

        H_2  =  3 x

Now since

     [NH_3] = 0.387  \  M

     x= 0.387  \  M    

H_2  =  3 * 0.173    

H_2  =  0.519 \ M    

NH_3 = 0.387 -2(0.173)

NH_3 = 0.041 \ M

Now the equilibrium constant is

           K_c  =  \frac{[N_2][H_2]^3}{[NH_3]^2}

substituting values

           K_c  =  \frac{(0.173) (0.519)^3}{(0.041)^2}

           K_c  = 14.39

         

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