Because its in group one....which only lose or gain electrons.....thus it will only have oneelectron to fulfill its octect rule
Both the increase in the boling point and the depression on the freezing point are colliative properties.
This is, they are proportional to the number of particles dissolved in the solvent, which is measured by the molality of the solution and the factor i (Van'f Hoff).
The answer to the question is that 1) the boling point of a solution of water and calcium chloride at standard pressure will be higher than the normal boiling point of pure water, and 2) the freezing point of a solution of water and calcium chloride at standard pressure will be lower than the normal freezing point of pure water.
Answer:
The famous oil drop experiment exploits that fact that an oil drop in an electronic field will get negative charge accumulation which can be balanced and observed in order to determine the charge of an electron.
Answer: Option (c) is the correct answer.
Explanation:
When an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.
This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. As a result, it helps in maintaining the pH of the solution.
In the give case, when we add the HCl then more number of protons will dissociate. This causes the acetate to react with the protons and leads to the formation of acetic acid.
We know that acetic acid is a weak acid and it dissociates partially or feebly. Therefore, no change in pH will take place.
Thus, we can conclude that equation 
represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.
Answer:
5.8 g
Explanation:
Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.
The amount of NaCl required is calculated as follows:
(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol
This amount is converted to grams using the molar mass (58 g/mol).
(0.1 mol)(58 g/mol) = 5.8 g
