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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Shortest side (a) = 58
middle side (b) = 64
longest side (c) = 77
the 3 sides a + b + c = 199
b = a + 6
c = a + 19
substitute your new values for b & c into your original formula, so:
a + (a+6) + (a+19) = 199
3a + 25 = 199
3a = 174
a = 58
then substitute 58 into your b & c formulas to figure out the rest
b = a + 6 = 58 + 6 = 64
c = a + 19 = 58 + 19 = 77
The rule is 3x
so to find the next number you multiple the most recent one by 3
the next three are:
32.4, 97.2, 291.6
Answer: No he would still have 4 leftovers
Step-by-step explanation: 64 divided by 6 equals 10 and the remainder of 4
