Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1 . Because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
Answer:
Explanation:
In the solution of AB , they are split to give ions as follows
AB ⇄ A⁺ + B⁻
Product of concentration of A⁺ and B⁻ in saturated solution of AB is constant .
This is called Ksp
Ksp = [A⁺] [ B⁻]
If product of concentration of A⁺ and B⁻ exceeds Ksp , the equilibrium shifts to the left side and excess ions come out of solution in the form of precipitate. So second option is the answer.
Answer:- 3333 g of solution.
Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.
(i) 0.045% NaCl by mass
Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?
We could solve this using proportions...
(0.045/100) = (1.5/X)
0.045(X) = 1.5(100)
0.045X = 150
X = 150/0.045 = 3333
So, 1.5 g of NaCl is present in 3333 g of solution.
Answer:
Yes
Explanation:
Is this a question or what?
It is either netforce or gforce
