Answer:
(b) Calculate the molarity of a solution of 4.8 mole of HCl in 600 mL of solution. ... (g) Calculate the mass of Na2CO3 that must be used to make 700 mL of a 0.136 M Na2CO3 ... (h) What mass of NaOH is needed to make 200 mL of a 0.300 M NaOH solution? ... However, when we are reacting solutions we have to convert.
Explanation:
<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>
Ph 2 will ahve more strength due to the fact that its more acidic compared to pH 5.
the lower the number of a pH, the more it is heading towards being acidic, but the higher the number, the more it heads towards being an alkali. here is a ppt i made along time ago. hope it can help you . have a nice day
Answer:
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