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Llana [10]
3 years ago
12

Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear rea

ctions we must account for protons, neutrons, and electrons, as well as write out the symbols for various chemical elements. In a nuclear equation, the products and reactants are symbolized as AZX where X is the chemical symbol for the element, A is the mass number, and Z is the atomic number. There are two main rules to remember when balancing nuclear equations:
The total of the superscripts (mass numbers, A) in the reactants and products must be the same.
The total of the subscripts (atomic numbers, Z) in the reactants and products must be the same

What is the value of A in the following nuclear reaction?

237Np93 →233 Pa91 +AZX
Chemistry
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

The correct answer is - 4.

Explanation:

As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -

The sum of the product mass number of products = mass of reactant

237Np93 →233 Pa91 +AZX is the equation,

Solution:

Mass of reactants = 237

Mass of products are - Pa =233 and A = ?

233 + A = 237

A = 237 - 233

A = 4

So the equation will be:

237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

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