Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
I believe is the answer is D. The block moves to the right. There is more force blowing to the right, so the block will move to the right.
Answer:
pKa of the acid is 3.6
Explanation:
When a weak acid, HX, reacts with NaOH, the conjugate base, X⁻, is produced:
HX + NaOH → X⁻ + Na⁺ + H₂O
At the half neutralized solution, [HX] = [X-]
Based on Henderson-Hasselbalch equation:
pH = pKa + log [ X⁻] / [HX]
<em>Where pH is the pH of the buffer = 3.6</em>
<em>pKa is the pka of the solution</em>
<em>And as [ X⁻] = [HX], [ X⁻] / [HX] = 1</em>
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Replacing:
3.6 = pKa + log 1
3.6 = pKa + 0
<h3>pKa of the acid is 3.6</h3>
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Answer:
2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom
Explanation:
Given electronic configurations are :
1st:
2nd :
given 1st ionization energy are: 2080 kJ/mol and 496 kJ/mol
generally ionization energy of fulfilled orbital is more than half filled orbital and these two state are more stable.
therefore ionization energy of fulfilled is more than half filled orbital
hence
ionization energy of 1st atom will be very high because its orbital is fulfilled and less energy for 2nd atom so 2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom.
The chemical reaction would be written as:
2Na + Cl2 = 2NaCl
Since we are given the amounts of the reactants available for reaction, we have to determine the limiting reactant. And use this amount to calculate for the theoretical yield.
55 g Na ( 1 mol / 22.99 g ) = 2.39 mol Na
67.2 f Cl2 ( 1 mol / 70.9 g ) = 0.95 mol Cl2
Therefore, the limiting reactant would be Cl2 since it is the one consumed completely.
0.95 Cl2 ( 1 mol NaCl / 1 mol Cl2) = 0.95 NaCl produced from the reaction