AgNO₃ will act as the oxidising agent.
<h3><u>For the given chemical equation:</u></h3>
Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂
Half reactions for the given chemical reaction:
<u>Reducing agent:</u>
Cu → Cu²⁺ + 2e⁻
Copper is a reducing agent because it is losing 2 electrons, which causes an oxidation process.
<u>Oxidising Agent</u>:
Ag⁺ + e⁻ → Ag
The silver ion undergoes a reduction process and is regarded as an oxidizing agent since it is acquiring one electron per atom.
Hence, AgNO₃ is considered as an oxidizing agent and therefore the correct answer is Option B.
<h3><u>
Oxidising and Reducing agents</u></h3>
- An oxidizing agent is a substance that reduces itself after oxidizing another material. It passes through a reduction process in which it obtains electrons and the substance's oxidation state is decreased.
- A reducing agent is a chemical that oxidizes after reducing another material. It passes through an oxidation process in which it loses electrons and the substance's oxidation state increases.
To know more about the process of Oxidation and Reduction, refer to:
brainly.com/question/4222605
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<span>B. muscular endurance
</span>The capacity of a muscle to continue contracting over a period of time without fatigue is called muscular endurance.
NOT:
A. muscular strength.
<span>C. cardiovascular endurance. </span>
<span>D. power.</span>
Answer:
So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.
Explanation:
What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.
Answer:
B. 126
Explanation:
the scale is 1 to 42 so the scale would be:
3 cm:x cm
3 * 42 = 126 cm