Answer:
Percentage composition = 14.583%
Explanation:
In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.
Percentage composition by mass of Nitrogen
Nitrogen = 14g/mol
In one mole of the compound;
Mass of Nitrogen = 1 mol * 14g/mol = 14g
Mass of compound = 1 mol * 96.0 g/mol = 96
Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100
percentage composition = 14/96 * 100
Percentage composition = 0.14583 * 100
Percentage composition = 14.583%
0.555 = 1.90/V
⇒ V = 1.90/0.555
⇒ V = 3.423 liter
<span><span>K_2</span>C<span>O_3</span>(aq)+Ca(N<span>O_3</span><span>)_2</span>(aq)→ ?</span>
If we break these two reactants up into their respective ions, we get...<span><span>
K^+ </span>+ C<span>O^2_3 </span>+ C<span>a^<span>2+ </span></span>+ N<span>O_−3</span></span>
If we combine the anion of one reactant with the cation of the other and vice-versa, we get...<span>
CaC<span>O_3 </span>+ KN<span>O_3</span></span>
Now we need to ask ourselves if either of these is soluble in water. Based on solubility rules, we know that all nitrates are soluble, so the potassium nitrate is. Alternatively, we know that all carbonates are insoluble except those of sodium, potassium, and ammonium; therefore, this calcium carbonate is insoluble.
This is good. It means we have a driving force for the reaction! That driving force is that a precipitate will form. In such a case, a precipitation reaction will occur, and the total equation will be...<span><span>
K_2</span>C<span>O_3</span>(aq) + Ca(N<span>O_3</span><span>)_2</span>(aq) → CaC<span>O_3</span>(s) + 2KN<span>O_3</span>(aq)</span>
To determine the net ionic equation, we need to remove all ions that appear on both sides of the equation in aqueous solution -- these ions are called spectator ions, and do not actually undergo any chemical reaction.
To determine the net ionic equation, let's first rewrite the equation in terms of ions...
2K^+(aq) + CO_3^{2-}(aq) + Ca^{2+}(aq) + 2NO_3^{-}(aq) → Ca^{2+}(s) + CO_3^{2-}(s) + 2K^+(aq) + 2NO_3^-(aq)
The species that appear in aqueous solution on both sides of the equation (spectator ions) are...
<span>
2K^+,NO_3^-</span>
If we remove these spectator ions from the total equation, we will get the net ionic equation...
CO_3^{2-}(aq) + Ca^{2+}(aq) <span>→</span> CaCO_3(s)
