<span>A- solute is the answer you were looking for
</span>
Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, 
= 102 pm
Radius of cation, 
= 196 pm
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
1. For the first question, we must find the mass of the anhydrous salt, MgSO₄. The molar mass for MgSO₄·7H₂O is 246.47 g/mol, while that of MgSO₄ is 120.37 g/mol.
7.834 g MgSO₄·7H₂O * 1 mol MgSO₄·7H₂O/246.47 g * 1 mol MgSO₄/1 mol MgSO₄·7H₂O * 20.37 g MgSO₄/mol = <em>0.647 g MgSO₄</em>
2. Mass of Water = Mass of sample - Mass of anhydrous sale
Mass of water = 7.834 - 0.647 = 7.19 g
Percent Water in Hydrate = 7.19/7.834 * 100 = <em>91.74%</em>
Compounds often differ from elements.
<span>Barium lies in the second group of the periodic table. Elements in this group have two valence shell electrons. These elements have a tendency to produce cations since it is easier to lose the two electrons than to gain six. Now, out of the options, chlorine forms an anion. It is an element from group 7, meaning it exhibits a valency of -1. Therefore, in order to bond with Barium, two Chlorine atoms will be needed. <u>The second option is correct.</u></span>
