<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
Answer:
4.6305 * 10^-6 mol^3.L^-3
Explanation:
Firstly, we write the value for the solubility of Ca(IO3)2 in pure water. This equals 0.0105mol/L.
We proceed to write the dissociation reaction equation for Ca(IO3)2
Ca(IO3)2(s) <——->Ca2+(aq) + 2IO3-(aq)
We set up an ICE table to calculate the Ksp. ICE stands for initial, change and equilibrium. Let the concentration of the Ca(IO3)2 be x. We write the values for the ICE table as follows:
Ca2+(aq). 2IO3-(aq)
I. 0. 0.
C. +x. +2x
E. x. 2x
The solubility product Ksp = [Ca2+][IO3-]^2
Ksp = x * (2x)^2
Ksp = 4x^3
Recall, the solubility value for Ca(IO3)2 in pure water is 0.0105mol/L
We substitute this value for x
Ksp = 4(0.0105)^2 = 4 * 0.000001157625 = 4.6305 * 10^-6
Answer:
the answer to ur question is B
Explanation:
heating curve- a graph / plot where a subject it increases in temperature against time to accurately measure it's amount of energy it absorbs and changes state with temperature that increase
it shows how temperature changes as a substance is heated up at a constant rate
Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Explanation:
Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L
Molarity = 0.750 M
It is known that molarity is the number of moles of solute present in liter of a solution.
Therefore, moles present in given solution are calculated as follows.

Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Answer:
positive(cations) and negative(anions)
Explanation: