Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
Ok so all are in the alkali family also they are all metals and in group 1 but I think the one that might reacts quickest is Li because it has less AMU
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
