If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?​

The question is incomplete, the complete question is;

AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)

1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

1.8 g

Explanation:

Given the equation;

AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)

We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).

If this is so,

Number of moles of precipitate formed = 4.52g/143.32 g/mol

Number of moles of precipitate formed = 0.0315 moles

Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl

Therefore;

Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

c, liquids evaporate, creating a gas