Calculate mass of Tollen’s Reagent Required?
18.70 kg
37.40 kg
9.35 kg
55.10 kg
The correct answer is option (A) : 18.70 kg
The balanced equation is :
No. of moles of NH3 formed = \(\frac{4\times 10^3}{17}\)
\(\therefore\) No. of moles of Tollen’s Reagent consumed = \(\frac{4\times 10^3}{17}\)
So, mass of Tollen’s Reagent = \(\frac{4\times 10^3}{17}\times159\)
= 18.70 kg
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
A coordination compound holds a central metal atom or ion surrounded by various oppositely charged ions or neutral molecules. These molecules or ions are re-bonded to the metal atom or ion by a coordinate bond.
A coordination entity composes of a central metal atom or ion bonded to a fixed number of ions or molecules.
A molecule, ion, or group which is bonded to the metal atom or ion in a complex or coordination compound by a coordinate bond is commonly called a ligand. It may be either neutral, positively, or negatively charged.