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.
The answer is (2) higher vapor pressure and weaker intermolecular forces. Propanone has a lower boiling point, so it is more volatile than water. Propanone's vapor pressure is, therefore, higher than that of water at 50 degrees Celsius. Propanone is more volatile due to the fact that the intermolecular forces that hold its molecuels together are not as strong as those that hold together molecules of water. Since the IMFs are weaker, it takes less thermal energy to break individual molecules free of each other.
A base is weak when only a little of it dissolved in Water.
It images help u...
Answer:
95.3g
Explanation:
Given parameters:
Mass of water = 540g
Percentage by mass of solution = 15%
Unknown:
Mass of sucrose = ?
Solution;
To solve this problem, we must understand that a solution is made up of solute dissolved in a solvent. A solute is the solid dissolved in another. The solvent is the is dissolving medium;
Percentage by mass of solute in solution = 
Mass of solution = mass of water + Mass of sucrose = 540 + M
where M is the mass of sucrose;
therefore;
= 15
Solving for M gives 95.3g
Mass of sucrose is 95.3g
