Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
Your answer would be, Gas atoms subjected to the electricity emit bright lines of light.
Hop that helps!!!
