Explanation:
Ksp of NiCO3 = 1.4 x 10^-7
Ksp of CuCO3 = 2.5 x 10^-10
Ionic equations:
NiCO3 --> Ni2+ + CO3^2-
CuCO3 --> Cu2+ + CO3^2-
[Cu2+][CO3^2-]/[Ni2+][CO3^2-]
= (2.5* 10^-10)/(1.4* 10^-7)
= 0.00179.
[Cu2+]/[Ni2+]
= 0.00179
= 0.00179*[Ni2+]
If all of Cu2+ is precipitated before Na2CO3 is added.
= 0.00179 * (0.25)
The amount of Cu2+ not precipitated = 0.000448 M
The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100
= 0.000448/0.25 * 100
= 0.18%
Therefore, percentage precipitated = 100 - 0.18
= 99.8%
The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.
The frequency of a 4,600 cm wave is 652 Hz
Frequency is the number of waves that pass a fixed point in unit time
Here given data is
Wave = 4,600 cm = 4.6 × 10⁸ nm
We have to calculated frequency = ?
Frequency =v = c/λ
Frequency = 3×10⁸m / 4.6 × 10⁸ nm
Frequency = 652 Hz
Know more about frequency
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<span>M(NO</span>₃<span>)</span>₂<span> fully separates into M</span>²⁺<span> and NO</span>₃<span> </span>²⁻<span> </span><span>and M(OH)</span>₂<span> partially separates
as <span>M</span></span>²⁺<span><span> and 2OH</span></span>⁻
<span>M(NO</span>₃<span>)</span>₂<span><span> </span>→
M</span>²⁺<span> + 2NO</span>₃²⁻
<span>0.202 M 0.202 M</span>
<span> M(OH)</span>₂<span>(s) ↔ <span>M</span></span>²⁺<span><span> (aq) + 2OH</span></span>⁻<span><span>(aq)</span></span>
<span>I - -</span>
<span>C -X +X +2X</span>
<span>E X 2X</span>
<span>Ksp = [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²
4.45 * 10∧-12 = (0.202
+ X ) (2X)²
Since X is very small, (0.202 + X ) = 0.202
<span>4.45 * 10<span>-12 </span>= 0.202 *
4X</span>²
<span> X = 2.347 </span>× 10∧-6 M
Hence
the solubility of <span>M(OH)2
is 2.347 </span>× 10∧-6 M
Answer:
there are no options but i would say Alaska
Explanation:
