Answer:
your answer would be the last one gold
Answer:
= 0.030 M
Explanation:
We can take x to be the concentration in mol/L of Ag2SO4 that dissolves
Therefore; concentration of Ag+ is 2x mol/L and that of SO4^2- x mol/L.
Ksp = 1.4 x 10^-5
Ksp = [Ag+]^2 [SO42-]
= (2x)^2(x)
= 4x^3
Thus;
4x^3 = 1.4 x 10^-5
= 0.015 M
molar solubility = 0.015 M
But;
[Ag+]= 2x
Hence; silver ion concentration is
= 2 x 0.015 M
= 0.030 M
Sodium Sulfate
= Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M
stands for Molarity, defined as Molarity = (moles of solute)/(Liters of
solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65
moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+
ions in solution for every one Na2SO4.
Therefore if
3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and
since this is still a theoretical solution, we can assume 1 L of solution.
Finally we find
[Na+] = 2*3.65 = 7.3M
Use the same
logic for parts b and c
Answer: if doing it on ck-12 the answer is 40.1
Explanation:
Answer:
5.7 moles of O2
Explanation:
We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:
2KClO3 —> 2KCl + 3O2
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.
Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.
Thus, 5.7 moles of O2 were obtained from the reaction.
