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The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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Answer:
C
Explanation:
This is essentially one of the several safety measures in the chemical laboratory. This particular approach is one used in the case of fire eventualities.
A is wrong
This is because in the advent of a fire incident, it is necessary to evacuate the building as a whole. Meeting in the hallway is still within the building which is not the right thing to do when there’s a fire outbreak. Occupants are expected to leave the building immediately
B. Is also wrong. Taking time to pack your belongings might make you be caught in the inferno. It is expected that you leave the building at once
The sample of smoke described above can be described as a heterogeneous mixture. This type of mixture do not have uniform properties and composition. So, getting a certain small sample would not represent the whole mixture since it does not have uniform composition.
Answer:
3.67 moles of N
Explanation:
The epinephrine's chemical formula is: C₉H₁₃O₃N
We were told that a chemist found that in a mesaure of epinephrine, he found 33 moles of C
We must know that 9 moles of C are in 1 mol of C₉H₁₃O₃N so, let's make a rule of three:
If 9 moles of C are found in 1 mol of C₉H₁₃O₃N
Therefore 33 moles of C must be found in (33 .1) / 9 = 3.67 moles of C₉H₁₃O₃N
There is a second rule of three, then.
In 1 mol of C₉H₁₃O₃N we have 1 mol of N
Then, 3.67 moles C₉H₁₃O₃N must have (3.67 . 1) / 1 = 3.67 moles of N
Remember 1 mol of C₉H₁₃O₃N has 9 moles of C, 13 moles of H, 3 moles of O and 1 mol of N
