What is the amount of diamine silver that can be formed when 10.00 g AgCl is mixed with 1.00 L of 0.100 M NH3?

The amount of diamine silver chloride = 8.87 g

Given

10 g AgCl

1.00 L of 0.100 M NH3

Required

the amount of diamine silver

Reaction

AgCl + 2 NH₃ → [Ag(NH₃)₂]Cl

mol AgCl :

= mass : MW

= 10 g : 143,32 g/mol

= 0.0698

mol NH₃ :

= M x V

= 0.1 x 1

= 0.1

NH₃ as a limiting reactant

mol [Ag(NH₃)₂]Cl based on NH₃ :

= 1/2 x mol NH₃

= 1/2 x 0.1

= 0.05

Mass diamine silver :

= 0.05 x  177.3822 g/mol

= 8.87 g

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.