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
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Answer:

Since the measurement can't be negative the correct answer for this case would be 
Step-by-step explanation:
Let's assume that the figure attached illustrate the situation.
For this case the we know that the original area given by:

And we know that the initial area is a half of the entire area in red
, so then:

And we know that the area for a rectangular pieces is the length multiplied by the width so we have this:

We multiply both terms using algebra and the distributive property and we got:

And we can rewrite the expression like this:

And we can solve this using the quadratic formula given by:

Where
if we replace we got:

And the two possible solutions are then:

Since the measurement can't be negative the correct answer for this case would be 
He paid him : 24t + 15
for 3 hrs...
24(3) + 15 = 72 + 15 = $ 87
Answer:
An initial amount of 180 grows at a rate of 22% every month.
=> The equation for the amount y after x months:
y = 180 x (1 + 22/100)^x
Hope this helps!
:)