<span>When one talks about ppm in a liquid solution someone means mg/L so we would not be using the density. This usually means ug/g or mg/kg
0.115 g Na^+ * 10^6 ug/1 g = 115000 ug/g
4.55 L * 1000 mL/1L = 4550 mL
Concentration of Na^+ in ppm:
115000 ug/g /4550 mL = 25.27 pm of sodium ion</span>
Answer:
-145.2kJ
Explanation:
Enthalpy is an extensive property as its value depends on the amount of substance present in the system.
If the enthalpy for one mole of methanol = -726 kJ/mol;
The Enthalpy for 0.2 mol is given as;
Enthalpy = 0.200 * 726
Enthalpy = -145.2kJ
It would take -145.2kJ for 0.200 mol of methanol to undego the combustion reaction.
<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
