In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide.
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂ - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present
so if 3 mol are present in 1000 ml
then volume for 0.038 mol = 1000/3 * 0.038
= 12.67 ml
Answer:
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Explanation:
Data
Aqueous acetic acid
Aqueous barium hydroxide
Formula
Acetic acid CH₃CH₂COOH
Barium hydroxide Ba(OH)₂
Neutralization reaction
CH₃CH₂COOH + Ba(OH)₂ ⇒ Ba( CH₃CH₂COO)₂ + H₂O
Barium acetate + H₂O
Barium acetate = Ba( CH₃CH₂COO)₂ = Ba(C₃H₅O₂)₂
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