Answer:
Six electrons are transferred in the formation of Al₂O₃.
Explanation:
Aluminium metal and Oxygen react to form Al₂O₃ as,
2 Al + 3/2 O₂ → Al₂O₃
Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms.
So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
