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Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:
Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:
Therefore, the maximum amount of nickel(II) cyanide is
Most transition metal form more than one cation but aluminum forms the Al3+ cation only.
Answer:
Reduction: 2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
Oxidation: H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
Explanation:
In H₂O₂, hydrogen has the oxidation number +1 and oxygen the oxidation number -1.
In the reduction half-reaction (H₂O₂ is the oxidizing agent), H₂O₂ forms H₂O. The oxidation number of oxygen decreases from -1 to -2.
2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
In the oxidation half-reduction (H₂O₂ is the reducing agent), H₂O₂ forms O₂. The oxidation number of oxygen increases from -1 to 0.
H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
2.45 °C
From the Ideal gas law (Combined gas law)
PV/T = P'V'/T' .....eq 1
Where:
P - initial pressure
V - initial volume
T - initial temperature
P' - final pressure
V' - final volume
T' - final temperature.
To proceed we have to make T the subject from eq.1
Which is, T = P'V'T/PV.......eq.2
We have been provided with;
Standard temperature and pressure (STP)
P = 760 mm Hg (SP - Standard Pressure)
T = 273.15 K (ST - Standard Temperature)
V = 62.65 L
P' = 612.0 mm Hg
V' = 78.31 L
T' = ? (what we require)
Therefore, we substitute the values into eq.2
T' =
.
T' = 275.60 K
T = (275.60 - 273.15) ......To °C
T = 2.45 °C
>>>>> Answer
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