A composition of reflections over parallel lines is the same as a __________.
2 answers:
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Complete Question
The complete question is shown on the first uploaded image
Answer:
X is Be
Y is B
Z is O
Explanation:
In order to understand the solution, we need to that the an element form a bond based on
1 The number of valence electron available in its atom
2 The number of electron it require to attains its stable state i.e its Octet state
The number of valence electron for each of the option are
F = 7, N = 5 , O = 6 , C = 4 , B = 3 , Be = 2
Considering the first element
We see that X is donating two valence electron in its atom and it is donating it to 2 Cl as Cl only requires a single electron to achieve its stable state
Now looking option we can see that Be has two valence electron in it atom and in order for it to attain its stable state it needs to release the electron so the correct option for this question is Be
Considering the second element
We see that Y is donating three valence electron in its atom and it is donating it to 3 Cl as Cl only requires a single electron to achieve its stable state
Now looking option we can see that B has three valence electron in it atom and in order for it to attain its stable state it needs to release the valence electron so the correct option for this question is B
Considering the second element
We see that Z is donating two valence electron in its atom and it is donating it to 2 Cl as Cl only requires a single electron to achieve its stable state, also Z has 2 lone-pair valence electron
Now looking at the option we can see that O has 6 valence electron so if it donates 2 valence electron it will still be left with 4 - pair of electron
Therefore the correct option to this question is O
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:
- Oxidation reaction
Li⁰(s) → Li⁺(aq) + e⁻ (2)
- Reduction reaction
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:
4.52 x 10²⁴ atoms Cl
Explanation:
A mole is a name that means a certain number like a dozen means 12. 1 mole of chlorine atoms is 6.022 x 10²³ chlorine atoms. The unit conversion is 6.022 x 10²³/mol.
Round to 4.52 x 10²⁴ atoms Cl for the correct number significant figures.