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1 answer:
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Answer:
5.63 g
Explanation:
Step 1: Write the balanced equation
CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)
Step 2: Calculate the reacting moles of copper (II) bromide
30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:
Step 3: Calculate the moles formed of silver (I) bromide
The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.
Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr
The molar mass of AgBr is 187.77 g/mol.
Answer:
See explanation and image attached for details
Explanation:
The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.
