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Natali5045456 [20]
3 years ago
8

1. Which of the following nuclear equations is correctly balanced? (choices attached)

Chemistry
1 answer:
lawyer [7]3 years ago
7 0

1. <em>Balancing nuclear equations </em>

Answer:

_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}

Explanation:

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. <em>Amount remaining </em>

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:  

  No. of                  Fraction       Amount  

<u>half-lives</u>   <u>t/(da</u>)  <u>remaining</u>  <u>remaining/g</u>  

      1             24            ½                21.0  

      2            48            ¼                10.5  

      3             72           ⅛                 5.25  

      4             96          ¹/₁₆                 2.62  

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.  

===============

3.<em> Calculating the half-life </em>

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min       Divide both sides by 2

  1 half-life = 5.2 min/2 = 2.6 min

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Complete Question

The complete question is shown on the first uploaded image

Answer:

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Explanation:

      From  the question we are told that

              The chemical reaction equation is

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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}

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          No \ of\ moles = \frac{3.54}{160}

                                = 0.022125 \ mols

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               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}

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            And      H_2  \ for  \ reactant

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                    K_c = \frac{0.022}{0.336}

                          K_c= 2.8*10^{-4}

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