Answer:
The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
a) To calculate the molarity of solution, we use the equation:

We are given:
Mass of solute (NaCl) = 1.46 g
Molar mass of sulfuric acid = 58.5 g/mol
Volume of solution = 

Putting values in above equation, we get:

0.09982 M is the concentration of the sodium chloride solution.
b) 
Moles of NaCl = 
according to reaction 1 mol of NaCl gives 1 mol of AgCl.
Then 0.02495 moles of NaCl will give:
of AgCl
Mass of 0.02495 moles of AgCl:

The mass of the precipitate that AgCl is 3.5803 g.
answer: dispersed from the liquid so cold air can take its place
Answer:
D because i did this last week and got it right.
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^4