The compound HClO4, when placed in water, will dissociate into the ions, H+ and ClO4-. Therefore, the 2.0 M solution will also form 2.0 M H+. The pH is calculated through the equation,
pH = -log[H+]
Substituting,
pH = -log[2] = -0.3
Thus, the pH of the solution is -0.3.
2SO₂ + O₂ = 2SO₃
n(O₂)=1 mol
n(SO₂)=2n(O₂)
n(SO₂)=2 mol
<span>N2 + 3H2 → 2 </span>NH3<span> from bal. rxn., 2 moles of </span>NH3<span> are formed per 3 moles of </span>H2, 2:3 moleH2<span>: 3.64 </span>g<span>/ 2 </span>g<span>/mole </span>H2<span>= 1.82 1.82 moles </span>H2<span> x 2/3 x 17
</span>
C6H6(l) + 6 Cl2(g) = C6Cl6(s) + 6 HCl(g)
Ans: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams