I hope I am right on this, but the answer is
Fe₂O₃ + 3CO --> 2Fe + 3CO₃
<span>c. Real gases and fusion reactions </span>
Answer:
<h2>Heterogeneous</h2>
Explanation:
<h3><em>Milk </em><em>seems</em><em> to</em><em> be</em><em> </em><em>homogeneous</em><em> mixture</em><em> </em><em>but </em><em>actually</em><em> </em><em>milk </em><em>is </em><em>a </em><em>heterogeneous</em><em> </em><em>mixture</em><em> </em><em>and </em><em>a </em><em>colloid</em><em> </em><em>solution</em><em>.</em></h3>
Food web should be the correct answer
:)
Answer:
mass H2O2 = 0.31g H2O2
mass unreacted = 0.0745g BaO2
Explanation:
Mw BaO2 = 169.33 g / mol
Mw HCl = 36.46 g/mol
Mw H2O2 = 34.0147 g/mol
LR:
⇒ mol BaO2 = 1.50g BaO2 * mol / 169.33 g = 8.86E-3 mol BaO2 / 1 = 8.86E-3 mol BaO2
⇒ mol HCl = 25.0 mL * 0.0272g / mL * mol / 36.46g = 0.0186 mol HCl / 2 = 9.3E-3 mol HCl.....L.R
mol H2O2:
⇒ 0.0186mol HCl * (mol H2O2 / 2mol HCl) = 9.325E-3 mol H2O2
⇒ g H2O2 = 9.325E-3mol H2O2 * 34.0147g H2O2 / mol H2O2 = 0.31g H2O2
mass of wich reagent is left unreacted:
mol that react BaO2 = 0.0186mol HCl * (mol BaO2 / 2mol HCl) = 9.3E-3mol BaO2
mol that unreacted BaO2 = 9.3E-3 - 8.86E-3 = 4.4E-4 mol BaO2
g unreacted BaO2 = 4.4E-4mol BaO2 * (169.33g BaO2 / mol BaO2) = 0.0745g BaO2
