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.
The answer would be E, it shows students how the different environments appear, since you can visually see it.
The answer is the fourth choice, or Have membrane-bound organelles. Hope this helps you!
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The answer is A) Scientific Law