Answer:
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Answer:
the land is heated through radiant energy,
Explanation:
Answer:
Reduction: 2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
Oxidation: H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
Explanation:
In H₂O₂, hydrogen has the oxidation number +1 and oxygen the oxidation number -1.
In the reduction half-reaction (H₂O₂ is the oxidizing agent), H₂O₂ forms H₂O. The oxidation number of oxygen decreases from -1 to -2.
2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
In the oxidation half-reduction (H₂O₂ is the reducing agent), H₂O₂ forms O₂. The oxidation number of oxygen increases from -1 to 0.
H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
Answer:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Explanation:
Both reactions are synthesis reactions (two substances combine to form another).
Reaction: K(s) + Cl₂(g) ⟶
The product is the binary salt KCl. The balanced chemical equation is:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
Reaction: Cu(s) + O₂(g) ⟶
The most likely product is the metal oxide CuO. The balanced chemical equation is:
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
