Answer:
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.
Explanation:
Answer:
1.75M
Explanation:
Given parameters:
Initial volume of acid, HCl = 450mL = 0.45L
Initial concentration = 3.5M
Final volume = 0.9L
Unknown:
Final concentration = ?
Solution:
This is a dilution problem in which a particular concentration is made from the stock of known concentration.
One important approach to solve this problem is to remember that the number of moles in the initial and final solution will always remain the same.
Since we know this;
Number of moles = molarity x volume;
let us find the number of moles of the initial solution;
Number of moles = 3.5 x 0.45 = 1.58moles
Now, to find the new molarity;
Molarity =
Input the parameters;
Molarity = = 1.75M
Answer:
i = 2.79
Explanation:
The excersise talks about the colligative property, freezing point depression.
Formula to calculate the freezing point of a solution is:
Freezing point of pure solvent - Freezing point of solution = m . Kf . i
Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)
48.1°C - 44°C = 0.15 m . 9.78°C/m . i
4.1°C / (0.15 m . 9.78°C/m) = i
i = 2.79
In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.
Answer:
The boiling point decreases as the volume decreases.
Explanation:
The Temperature - Volume law otherwise called as Charles law is applied, which says that the volume of the given gas at constant pressure is directly proportional to the temperature measured in Kelvin. As the volume increases, the temperature also increases, if the volume decreases, then the temperature also decreases.
As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.