Answer:
makes nice
Explanation:
he really helps me somtimes
Answer:
solubility in presence of 0.16M Cu(IO₃⁻)₂ = 3.4 x 10⁻⁴M*
Explanation:
Cu(IO₃⁻)₂ ⇄ Cu⁺² + 2(IO₃⁻)
C(i) ---------- 0.16M 0M
ΔC ---------- +x +2x
C(f) ---------- 0.16 + x ≅ 0.16M* 2x
Ksp = [Cu⁺²][IO₃⁻]²
7.4 x 10⁻⁸M³ = 0.16M(2x)² = 0.64x²
x = solubility in presence of 0.16M Cu(IO₃⁻)₂ = SqrRt(7.4x10⁻⁸M³/0.64M²)
= 3.4 x 10⁻⁴M*
*Note: This is consistent with the common ion effect in that a reduction in solubility is expected. The normal solubility of Cu(IO₃⁻)₂ in pure water at 25°C is ~2.7 x 10⁻³M.
Answer: hello your question is incomplete below is the complete question
Salt water contains n sodium ions (Na+) per cubic meter and n chloride ions (Cl−) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt water. The cross sectional area of the pipe is A. The magnitude of the drift velocity of the sodium ions is VNa and the magnitude of the drift velocity of the chloride ions is VCl.
What is the magnitude of the ammeter reading ?
answer :
| I | = neAVₙₐ + neAV (Cl-)
Explanation:
Given that there are N sodium ions
<u>Determine the Magnitude of the ammeter reading </u>
| I | = current due to sodium ions + current due to (Cl-) ions
= neAVₙₐ + neAV (Cl-)
Molar mass CuNO₃ = 125.55 g/mol
number of moles:
4.80 / 125.55 => 0.038 moles of CuNO₃
M = n / V
0.270 = 0.038 / V
V = 0.038 / 0.270
V = 0.1407 L
hope this helps!
