Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
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To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.
Then knowing volume of aluminum and the density of it, we can solve for the mass.
D = m/v
Dv = m
2.7 g/ml • 8 ml = 21.6 grams.
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
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