Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Eight because an octet has 8 electrons and when it’s full it’s stable. If it’s not full, it’s constantly sharing or borrowing electrons.
Answer:
0.5
Explanation:
1 mole of ammonium nitrate contains 2 moles of nirogen
1 mole of nitrogen converts to 0.5 moles of ammonium nitrate
the conversation factor is 0.5
Answer:
90.99 or 91.0
Explanation:
Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.
