Based on the elements and charges in Copper (II) Oxalate, CuC₂O₄(s), the solubility in pure water is 1.7 x 10⁻⁴ M.
<h3>What is the solubility of Copper (II) Oxalate in pure water?</h3>
The solubility equilibrium (Ksp) is 2.9 x 10⁻⁸ so the solubility can be found as:
Ksp = [Cu²⁺] [C₂O₄²⁻]
Solving gives:
2.9 x 10⁻⁸ = S x S
S² = 2.9 x 10⁻⁸
S = 1.7 x 10⁻⁴ M
Find out more on solubility at brainly.com/question/23659342.
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im pretty sure the answer is 0.7
Answer:
This is because it is easier this way.
You avoid the problem of putting an element in one set and then realizing that the element also belongs to another set, then you must erase something in order to put the element in both sets. Starting with the intersection allows you to know where will be each set in the diagram, and in this way, the diagram should end up being more readable or "clean".
This may seem small, but being "clean" when doing math, will allow you to have an easier time dealing with a lot of problems. And also will be easier for other people when reading your equations and such.
Answer:
x= -23
Step-by-step explanation
Subtract 8 from itself and form -15
x+8= -15
-8 -8
x= -23
