How many moles of precipitate will be formed when 100.0 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction? 2 NaBr (aq) + Pb(NO₃)₂ (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)

Answer : The number of moles of precipitate, formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

Now we have to calculate the moles of precipitate, formed.

The balanced chemical reaction is:

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of

So, 0.02 moles of NaBr react to give mole of

Therefore, the number of moles of precipitate, formed will be 0.01 moles.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.