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vivado [14]
3 years ago
7

Pls help! Describe ionic and covalent bonding. Give an example of each

Chemistry
1 answer:
ankoles [38]3 years ago
6 0

Answer:

Example of ionic bonds— Sodium chloride and calcium oxide. Example of covalent bond— Water, carbon

Explanation:

ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound. Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.

A covalent bond consists of the mutual sharing of one or more pairs of electrons between two atoms. These electrons are simultaneously attracted by the two atomic nuclei. A covalent bond forms when the difference between the electronegativities of two atoms is too small for an electron transfer to occur to form ions.

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Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
olga nikolaevna [1]

Explanation:

The reaction equation for the given reaction will be as follows.

               HCl + NaOH \rightarrow NaCl + H_{2}O

Each mole of both HCl and NaOH gives one mole of water.

Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol dm^{-3} of NaOH (HCl).

It is known that 1 cm^{3} = 0.001 liter. So, 87 cm^{3} = 0.087 liter.

Hence, number of moles of water obtained from the given reaction are as follows.

                  0.087 liter × 1.6 mol = 0.1392 moles

            No. of moles = \frac{mass}{molar mass of water}

                0.1392 moles = \frac{mass}{18 g/mol}

                       mass = 2.5056 g

Now, volume of water present before the reaction is 2 \times 0.087 liter = 0.174 liter or 0.174 Kg (as density is 1 Kg/cm^{3}) or 174 g (as 1 kg = 1000 g).

Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g

Formula to calculate enthalpy of neutralization is as follows.

                Enthalpy of neutralization = mS \Delta T

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where,            m = mass

                       S = specific heat capacity

                   \Delta T = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = mS \Delta T

                     = 176.5056 g \times 4.18 J/K g \times (317.4 K - 298 K)              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.

5 0
4 years ago
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