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
Answer:
1 - e, 2 - k, 3 - a, 4 - i, 5 - b,
Explanation:
The ratio of the amount of analyte in the stationary phase to the amount in the mobile phase. --- Retention factor.
Time it takes after sample injection into the column for the analyte peak to appear as it exits the column. -- Retention time
The process of extracting a component that is adsorbed to a given material by use of an appropriate solvent system. -- Elution
Measure of chromatographic column efficiency. The greater its value, the more efficient the column. -- Theoretical plate number
Gas, liquid, or supercritical fluid used to transport the sample in chromatographic separations. -- Mobile phase
Immiscible and immobile, it is packed within a column or coated on a solid surface. -- Stationary phase
Answer:
52.2 g
Explanation:
Step 1: Write the balanced equation
3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O
Step 2: Calculate the moles corresponding to 89.7 g of KOH
The molar mass of KOH is 56.11 g/mol.
89.7 g × 1 mol/56.11 g = 1.60 mol
Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH
The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.
Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.533 mol × 97.99 g/mol = 52.2 g
The heat is involved in the production of 5.0 mol of MgO is 180 Kj
calculation
2 Mg (s) +O2 → 2 MgO
From the equation above two moles of MgO is used therefore
72.3 is for 2 moles
that is 72.3 kj = 2moles
? = 5 moles
by cross multiplication
= 72.3 kj × 5/2 = 180 Kj
