it would distract the pain
Answer:
93.17 g
Explanation:
Recall that: mol = mass/molar mass
Also; mol = molarity x volume
mole of (NH4)2SO4 to be prepared = 282/1000 (dm3) x 2.50 (mol/dm3)
= 0.705 mol
This can be used to determine the mass of (NH4)2SO4 that will be required.
mass = mole x molar mass
Hence, mass of (NH4)2SO4 required = 0.705 x 132.15
= 93.17 g
Hence, the mass of ammonium sulfate that will be required to prepare 282 mL of a 2.50M solution is 93.17 g
2. Bio Chemistry
3. Organic Chemistry
4. Inorganic Chemistry
Answer:
Organic chemistry
Explanation:
Answer: Option (B) is the correct answer.
Explanation:
A solution that has maximum concentration of solute particles and on adding more solute the particles of solute remains undissolved is known as a saturated solution.
Whereas when particles of solute keep on dissolving in a solution then it is known as an unsaturated solution.
Thus, we can conclude that a solute is added to water and a portion of the solute remains undissolved. When equilibrium between the dissolved and undissolved solute is reached, the solution must be saturated.
(2) higher
(3) the same
Answer : The correct option is, (1) lower
Explanation :
Formula used for lowering in freezing point :
where,
= change in freezing point
= freezing point constant
m = molality
i = Van't Hoff factor
According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.
Now we have to calculate the Van't Hoff factor for the given solutions.
(a) The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2
(b) The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3
The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.
Thus, compared to the freezing point of 1.0 M at standard pressure, the freezing point of 1.0 M at standard pressure is lower.
Answer : The net ionic equation will be,
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The balanced ionic equation will be,
The ionic equation in separated aqueous solution will be,
In this equation, are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,