Which of the following explains why the entropy change is greater for the dissolution of nai compared to the dissolution of nabr?

a. iodide has weaker ion-dipole interactions with water than bromide.
b. the interactions between bromide ions with other bromide ions is stronger than the interactions between iodide ions with other iodide ions.
c. the more negative change in enthalpy observed with nai implies greater dissociation and hence greater entropy.
d. the bromide ion has more negative charge than the iodide ion. therefore, because of the greater charge, it forms a stronger ion-dipole network with water.
e. the cation forms stronger ion-dipole networks with water in nabr than nai because of the weaker bond to br.

Iodide has weaker ion-dipole interactions with water than bromide.

Explanation:

The electronegativity difference between sodium and iodine in sodium iodide is about 1.73. This shows that the compound is not composed of purely ionic bonds. Electro negativity decreases down the group hence iodine is far less electronegative than bromine and is thus ineffective in forming strong dipole interactions with water hence the higher entropy due to much less association of ions in solution.

hi,

the capital city of australia is canberra.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"