A student combines a solution of NaCl(aq) with a solution of AgNO3(aq), and a precipitate forms. Assume that 50.0mL of 1.0MNaCl(aq) and 50.0mL of 1.0MAgNO3(aq) were combined. According to the balanced equation, if 50.0mL of 2.0MNaCl(aq) and 50.0mL of 1.0MAgNO3(aq) were combined, the amount of precipitate formed would:

Based on the given information, the balanced equation is:

NaCl + AgNO3 = AgCl + NaNO3

Now the moles present in 50 ml of 1M NaCl is,

= 50 * 1/1000 mole = 0.05 mole

And the moles present in 50 ml of 1 M AgNO3 is,

= 50*1/1000 mole = 0.05 mole

Therefore, 0.05 mole of NaCl combines with 0.05 mole of AgNO3 to precipitate 0.05 mole of AgCl.

Now in the second case, to balance the chemical equation, 50 ml of 2 M NaCl combines with 50 ml of 1 M AgNO3. So, the moles present in 50 ml of 2 M NaCl will be,

= 50*2/1000 mole = 0.1 mole

However, the amount of AgNO3 in the second case is not changing, therefore, the amount of AgCl precipitated will be,

= 0.1 - 0.05 = 0.05 mole

Therefore, the amount of precipitate would not change as there is no change in the amount of AgNO3.

hloride, what is the mass of calcium in

explanation: