Answer:
5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.
Explanation:
Both fluorine and carbon are located in the second period of the periodic table, it means that they have 2 shells, so the valence electrons are found at about the same distance from their respective nuclei.
But fluorine has a higher atomic number, 9, than the carbon, 6. The atomic number represents how many protons there are in the nucleus, then there are more protons (positive charge) at the fluorine nucleus, and because of that, the attraction force between the nucleus and the valence electron is stronger in fluorine.
If the force is stronger, it will be necessary more energy to break the bond, so it will be harder to remove an electron from fluorine than from carbon.
The product formed when benzene reacts with isobutyl chloride in the presence of AlCl3 is tertiary butyl benzene.
THE CHEMICAL REACTION AS
step 1 : CH₃ - CH - CH₂Cl + AlCl₃ → CH₃ - CH₂ -CH₂ - CH₂⁺ + AlCl⁻
step 2: CH₃ - CH₂ - CH₂ - CH₂⁺ ---H SHIFT--→ (CH₃)₃C⁺
PRODUCT
C₆H₆ + (CH₃)₃ C⁺ → C₆H₆ ( CH₃)₃ C⁺
Hence the product formed is tertiary butyl benzene.
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Answer: gravity affect the amount of both kinetic and potential energy
Explanation:
Answer:
3.918 mol Al
Explanation:
To convert between moles and grams, you have to use the molar mass of the substance. The molar mass of aluminum is 26.98 g/mol. You use this as the unit converter.
Round the number to the lowest number of significant figures; 3.918 mol Al
Answer:
E° = 0.00 V
E = 0.079 V
Explanation:
We can identify both half-reactions occurring in a concentration cell.
Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻ E°red = -1.66 V
Cathode (reduction): Al³⁺(0.100 M) + 3 e⁻ → Al(s) E°red = -1.66 V
The global reaction is:
Al(s) + Al³⁺(0.100 M) → Al³⁺(1.0 × 10⁻⁵ M) + Al(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = -1.66 V - (-1.66 V) = 0.00 V
To calculate the cell potential (E) we have to use the Nernst equation.
E = E° - (0.05916/n) .log Q
where,
n: moles of electrons transferred
Q: reaction quotient
E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)
E = 0.079 V
