Answer:
The answer to your question is A.
Pure substances can not be broken down into others, so they cannot be molecules
Explanation:
Answer:
The answer to your question is P2 = 2676.6 kPa
Explanation:
Data
Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml
Temperature 1 = T1 = -108°C Temperature 2 = 22°C
Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?
Process
- To solve this problem use the Combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
- Convert temperature to °K
T1 = -108 + 273 = 165°K
T2 = 22 + 273 = 295°K
- Convert volume 2 to liters
1000 ml -------------------- 1 l
855 ml -------------------- x
x = (855 x 1) / 1000
x = 0.855 l
-Substitution
P2 = (12.8 x 100 x 295) / (165 x 0.855)
-Simplification
P2 = 377600 / 141.075
-Result
P2 = 2676.6 kPa
The lower the activation energy required for a chemical reaction
What's the problem ? Hardness is not the definition of a metal.
You need to expand your thinking. EVERY element is solid, liquid, and gas, over different ranges of temperature ... including all of the metals. There are only TWO elements that are liquid AT ROOM TEMPERATURE, and mercury is one of them. But on a mild day at the south pole, mercury is solid too.
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!