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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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179.1 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.1oC. After 306.9 g
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Answer:

the specific heat of the unknown compound is c_u=0.412J/g \cdot C

Explanation:

Generally the change in temperature of water is evaluated as

                \Delta T = T_2 -T_1

Substituting 16.1°C for T_1 and 27.4°C for T_2

                \Delta T = 27.4 - 16.1

                       =11.3^oC

Generally the change in temperature of  unknown compound is evaluated as

                  \Delta T_u = T_3 -T_2

Substituting 27.4°C for T_2 and 94.3°C for T_3

                                    \Delta T = 94.3 - 27.4

                                           =66.9^oC

Since there is an increase in temperature then heat is gained by water and this can be evaluated as

               H_w = mc_w \Delta T

Substituting 179.1 g  for m , 4.18 J/g.C for c_w(specific heat of water)

             H_w = 4.18 * 179.1 * 11.3

                   = 8459.6J

Since there is a decrease in temperature then heat is lost by unknown compound and this can be evaluated as

                    H_u = m_uc_u \Delta T_u

By conservation of energy law

       Heat lost  = Heat gained  

Substituting 306.9 g  for m_u , 8459.6J for H_u

           8459.6 = 306.9 * c_u * 66.9

  Therefore     c_u = \frac{8459.6}{308.9 *66.9}

                           =0.412J/g \cdot C

                   

4 0
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Moles/MxL 

1.709moles/2.10L =<span>0.814M</span>
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