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Mars2501 [29]
3 years ago
13

What is the charge of an atom after it gains an electron during the formation of a bond?

Chemistry
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

<em>D. One negative charge</em>

Explanation:

During the formation of a bond, if an atom gains an electron, after that it will be left with a negative charge compared to the atom before the bond is formed. This is because in these types of bonds, which are <em>ionic bonds</em>, there is a <em>transfer of electrons between atoms</em>, there will be one or more atoms that yield electrons that will be captured by another and other atoms that gain them, and the difference of charges produced by this transfer of electrons, will cause the union to occur due to the attraction between electrostatic forces.

If you have a neutral atom before joining, and it gains an electron to form a bond,<em> it will have one electron more than its initial state</em> (in the initial state, the number of protons and electrons is the same, because the atoms they are electrically neutral), so having an extra electron will make it have a negative charge, since there will be a difference between the number of protons and electrons that the atom possesses. <em>This is why the correct answer is D. </em>

In the case of <em>response A and B</em>, <em>the atom could only remain positively charged if it loses electrons</em>, but as in this case it wins, <em>they are not correct</em>.

<em>The answer C is also not correct</em> because only one electron wins, so that it is left with two negative charges, <em>it should gain two electrons during the bond formation.</em>

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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