The most abundant element in the Sun and in the stars are hydrogen and helium. Like most of the stars, there is a spontaneous radioactive reaction happening in the Sun. Hydrogen is transformed into Helium. As long as the stars are young, the most abundant element is hydrogen.
Answer:
1 mole of P₄ consumed:4 moles of PBr₃ produced
Explanation:
Look at the coefficients in the balanced chemical equation to determine the ratios. For every 1 mole of P₄ consumed (P₄ has a coefficient of 1), 4 moles of PBr₃ would be produced (PBr₃ has a coefficient of 4).
1. estuary
2. shallow waters of the caribbean sea
<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)
