Answer:
5.41 g
Explanation:
Considering:
Or,
Given :
For tetraphenyl phosphonium chloride :
Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)
Volume = 0.45 L
Thus, moles of tetraphenyl phosphonium chloride :
Moles of TPPCl = 0.01485 moles
Molar mass of TPPCl = 342.39 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of TPPCl = 5.0845 g
Also,
TPPCl is 94.0 % pure.
It means that 94.0 g is present in 100 g of powder
5.0845 g is present in 5.41 g of the powder.
<u>Answer - 5.41 g</u>
I believe they’re both true.
Answer:
Explanation:
The solution has been attached
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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The slight positive charges on the hydrogen atoms in water molecules attract the slight negative charges on the oxygen atoms of the other water molecules