PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN
NH₄⁺ + H₂O -----> H₃O⁺ + NH₃
CN⁻ + H₂O -----> HCN + OH⁻
Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:
Ka(HCN) = 4.9 x × 10⁻¹⁰
Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴
Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw
</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰
Then, Kb for CN⁻:
Kb(CN⁻) x Ka(HCN) = pKw
Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.
Answer:
Mass = 135.66 ×10⁻²¹ g
Explanation:
Given data:
Number of molecules of CuSO₄= 5.119×10²
Mass of CuSO₄= ?
Solution:
The given problem will solve by using Avogadro number.
1 mole contain 6.022×10²³ molecules
5.119×10² molecules ×1 mol / 6.022×10²³ molecules
0.85×10⁻²¹ mol
Mass in grams:
Mass = number of moles × molar mass
Mass = 0.85×10⁻²¹ mol × 159.6 g/mol
Mass = 135.66 ×10⁻²¹ g
Answer: 2.8 moles
Explanation:
The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.
Na2O + H2O --> 2NaOH
1 mole of H2O = 2 moles of NaOH
Let Z moles of H2O = 5.6 mole of NaOH
To get the value of Z, cross multiply
5.6 moles x 1 mole= Z x 2 moles
5.6 = 2Z
Divide both sides by 2
5.6/2 = 2Z/2
2.8 = Z
Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH
There would be 67 left because you do
Answer:
a). Coordination no. of
= 6
b). Coordination no. of
= 6
Explanation:
Coordination number is defined as number of donar atoms bonded to the central atom of the complex ion.
a). Coordination no. of
= 6
en or ethylenediamine is a bidentate ligand.
In bidentate ligand, two atoms directly bonded to the central atom.
NH3 is a unidentate ligand.
So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand
= 
b). Coordination no. of
= 6
Ethylenediamine (en) is a bidentate ligand.
oxalate ion (ox) is also a bidentate ligand.
Cl is a unidentate ligand
So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand
= 