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:
4
Step-by-step explanation:
4×(1/4) = 1
⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄
(x)(3x - 5) = 12
3x^2 - 5x = 12
3x^2 - 5x - 12 = 0
(x - 3)(3x + 4) = 0
x = 3, -4/3
The answer must be negative so the number is -4/3.
Hope this helps!
The answer is 0.3 (1/3.)
Proof?
f(0.3) = 18(0.3) + 8
f(0.3) = 6 + 8
f(0.3) = 14
14 = 14
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