Answer:
<span>Chlorine (Cl) is the oxidizing agent because it gains an electron.
Explanation:
Reaction is as follow,
</span><span> Cl</span>₂<span> (aq) + 2 Br</span>⁻<span> (aq) </span>→ <span> 2Cl(aq) + Br</span>₂ <span>(aq)
Oxidation Reaction:
2 Br</span>⁻ → Br₂ + 2 e⁻
Two atoms of Br⁻ (Bromide) looses two electrons to form Br₂ molecule. Hence it is oxidized and is acting as reducing agent.
Reduction Reaction:
Cl₂ + 2 e⁻ → 2 Cl⁻
One molecule of Cl₂ gains two electrons to form two chloride ions (Cl⁻). Therefore, it is reduced and has oxidized Br⁻, Hence, acting as a oxidizing agent.
Answer: E
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The number of mole of O₂ produced from the decomposition of H₂O₂ is 0.0033 mole
<h3>How to determine the number of mole of O₂ produced</h3>
We'll begin by obtaining the mole of H₂O₂ that decomposed. This is obatined as follow:
- Pressure (P) = 0.978 atm
- Temperature (T) = 27 °C = 27 + 273 = 300 K
- Volume (V) = 0.082 L
- Gas constant (R) = 0.0821 atm.L/Kmol
- Mole of H₂O₂ (n) =?
PV = nRT
Divide both sides by RT
n = PV / RT
n = (0.978 × 0.082) / (0.0821 × 300)
n = 0.0033 mole
Finally, we shall determine the number of mole of O₂ produced. This is illustrated below
H₂O₂ -> H₂ + O₂
From the balanced equation above,
1 mole of H₂O₂ decomposed to produce 1 moles of O₂
Therefore,
0.0033 mole of H₂O₂ will also decompose to produce = 0.0033 mole of O₂
Thus, number of mole of O₂ produced from the reaction is 0.0033 mole
Learn more about number of mole:
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Answer:
C
Explanation:
The Law of Conservation of Mass states that matter cannot be created or destroyed and must be conserved. I is stating matter cannot be created, which is true. II is saying that matter can only be rearranged/conserved, which again is true. III states matter cannot be destroyed, which is also true. IV is incorrect because it doesn't make sense. Matter does matter.
Explanation:
The given data is as follows.
Cross-section area of the rectangular is as follows.
Area = 
= 
= 
Tension applied on the specimen is 35,500 N
Now, formula to calculate the modulus of elasticity for a material of aluminium is as follows.
E = 70 GPa
= 
or, = 
Now, stress on the specimen is as follows.

= 
= 
According to Hook's law, we will calculate the resulting stain as follows.


= 
=
Thus, we can conclude that the resulting strain is
.