Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
The answer is a. Chemical change.
Answer:
27.9 g
Explanation:
CsF + XeF₆ → CsXeF₇
First we <u>convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles</u>, using its<em> molar mass</em>:
- Molar mass of CsXeF₇ = 397.193 g/mol
- 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇
As <em>1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇</em>, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.
Now we <u>convert 0.184 moles of CsF to moles</u>, using the <em>molar mass of CsF</em>:
- Molar mass of CsF = 151.9 g/mol
- 0.184 mol * 151.9 g/mol = 27.9 g
A is the answer to the question you asked.
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