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:
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Filtration<span>, Reabsorption, Secretion: The Three </span>Steps<span> of Urine Formation. The kidneys </span>filter<span> unwanted substances from the blood and produce urine to excrete them. There are three main </span>steps<span> of urine formation: glomerular </span>filtration<span>, reabsorption, and secretion.</span>
Potential energy<span> is the stored </span>energy<span> in an object due of its position or its configuration where as </span>Kinetic energy<span> is the </span>energy<span> which a body possesses because of its motion.</span>
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
