The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Superscript 100 Subscript 40 Baseline Z r + blank + 2 Superscript 1 Subscript 0 Baseline n What particle completes the equation? Superscript 138 Subscript 54 Baseline X e Superscript 139 Subscript 54 Baseline X e Superscript 119 Subscript 54 Baseline X e Superscript 138 Subscript 52 Baseline T e

The correct option is the first option

Explanation:

The equation described in the question is shown below

²³⁹₉₄Pu +¹₀n ⇒ ¹⁰⁰₄₀Zr + blank + 2¹₀n

This equation is a nuclear fission because it involves the splitting of a heavy nucleus, Plutonium (Pu), into smaller nuclei, Zirconium (Zr) and an unknown nuclei.

The law of conservation of matter states that matter can neither be created nor destroyed hence in other to get the missing atom, we must know the total number of subscripts (mass number) and superscripts (atomic number) on both sides.

The total mass number on the reactant side is 239 + 1 = 240

The total atomic number on the reactant side is 94 + 0 = 94

While, The total mass number on the product side is 100 + 2(1) = 102

The total atomic number on the product side is 40 + 2(0) = 40

To determine the missing atom, you subtract the the total atomic number of the product from that on the reactant side; 94 - 40 = 54

The atom with the atomic number 54 is Xenon,

To determine the mass number of the xenon isotope involved, you subtract the the total mass number of the product from that on the reactant side; 240 - 102 = 138.

Hence, the particle that completes the equation is ¹³⁸₅₄Xe, the first option

A on edge

Explanation:

the person above me explained it well but i wanted to shorten it.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole