Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.
Explanation:
As the charge of all electrons are equal, the repulsive force exerted by each of them is also going to be equal. So, as K has more electrons repulsing its valence electron than Na, it has greater electron shielding.