The answer is dipole-dipole and dipole-induced dipole forces.
The dipole-induced dipole is a kind of interaction induced by a polar molecule by disturbing the arrangement of electrons.
- In methyl cyclohexanone molecules, there is a permanent dipole moment due to dipole moment vectors not canceling.
- There is induction of dipole by disturbing the electronic arrangement.
- A permanent dipole moment is created in this interaction.
- Dipole-dipole interactions are defined as the forces that is formed from the close linkage of permanent or induced dipoles.
- These forces are called Van der Waal forces.
- Proteins contain a large number of these interactions, which vary considerably in strength.
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1) Only if new observations led to a change in our current knowledge.
2) <span>place their own moral views about the needs of society
3) FALSE
4) FALSE
5) FALSE(i think)
6) TRUE
7) Distillation
8) the # of neutrons
9) the element becomes a different element
10) atomic radii increase
11) decreases
i hope these are right and helps. please give me brainliest</span>
Answer:
E) 2.38
Explanation:
The pH of any solution , helps to determine the acidic strength of the solution ,
i.e. ,
- Lower the value of pH , higher is its acidic strength
and ,
- Higher the value of pH , lower is its acidic strength .
pH is given as the negative log of the concentration of H⁺ ions ,
hence ,
pH = - log H⁺
From the question ,
the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,
Therefore , the concentration of H⁺ = 0.0042 M .
Hence , using the above equation , the value of pH can be calculated as follows -
pH = - log H⁺
pH = - log ( 0.0042 M )
pH = 2.38 .
Answer is: (2) Chemical energy is converted to electrical energy.
An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.
In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).
Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.
Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).
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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