Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).
- (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:
- (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):
- (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:
%error=
%error=
%error=2.44 %
I think the answer is codominant??
37 grams of NaCl (when I mean equivalent I mean the ratio of the equation is 1:2 for moles or Cl2 and NaCl
Answer:
Db
Dubnium/Symbol
Explanation:
please mark my answer in brainlist plz
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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I hope it helps you!
