-13m = -377
m= -377/-13
m=29
the negatives cancel out. so just do 377/13 which equals 29
The equation that would represent the new area N of the floor of the cage would be
N=w2+4w which would be a
Your answer is C. 2000 ft^2/h.
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:
your answer is 0 because it's a horizontal line