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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Hey there!
Volume in mL :
1.68 L * 1000 => 1680 mL
Density = 0.921 g/mL
Therefore:
Mass = density * Volume
Mass = 0.921 * 1680
Mass = 1547.28 g
<u>Ans: 650 J = 155 calories</u>
<u>Given:</u>
Energy in joules = 650 J
<u>To determine:</u>
The energy in calories
<u>Explanation:</u>
1 joule = 0.2388 calories
Therefore, 650 joules = 0.2388 calories * 650 J/1 J = 155 calories
Your question isn't quite clear, but if you're wondering if a chemical is polar or non-polar, you simply draw a VSEPR sketch and draw arrows where the bonds are. Only draw arrows between atoms, NOT between an atom and a lone pair of electrons. The arrow should point to the most electronegative atom (you should be given an electronegativity scale). Afterwards, you add up the arrows as vectors, and look at the sum of the vectors. If the sum is zero (CH4 is a good example), the chemical is non-polar. If the sum is a vector, the chemical is polar (H2O, or water, is polar).
