1. 25 ml
2. 0.56 M
3. 3.0x 10^-2 M
Step by step explanation as follow
Use the formula (Molarity x volume) = ( molarity x volume)
To solve for the solubility of this compound:Cu ^ + 2 + CO3 ^ -2 = [x] [x] = x^2 1.4 x 10 ^ -10 = x^2 x = 1.1832 x 10 ^ -5 1.1832 x 10 ^ -5 moles/ L x 123.526 g/mol (molar mass) is equals to 1.5 x 10 ^-3 g/L is the solubility of this compound - CuCO3.
<span>We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:
31.3 g </span>(NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol <span>ammonium ions</span>
Answer:
ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)
Explanation:
The reactions are:
(1) ClO⁻(aq) + H₂O(l) → HClO(aq) + OH⁻(aq) [fast]
(2) I⁻(aq) + HClO(aq) → HIO(aq) + Cl⁻(aq) [slow]
(3) OH⁻(aq) + HIO(aq) → H₂O(l) + IO⁻(aq) [fast]
By adding up the 3 equations we get:
ClO⁻(aq) + <em>H₂O(l)</em> + I⁻(aq) + <em>HClO(aq)</em> + <em>OH⁻(aq)</em> + <em>HIO(aq)</em> → <em>HClO(aq)</em> + <em>OH⁻(aq)</em> + <em>HIO(aq)</em> + Cl⁻(aq) + <em>H₂O(l)</em> + IO⁻(aq)
And by canceling common terms on both sides, we can get the overall equation:
ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)
Therefore, the overall equation is ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq).
I hope it helps you!
