The acid having the yellow anion is a weak acid.
The weak acid is the acid that does not dissociate completely in solution. Strong acids are known to dissociate completely in solution. Hence, their cations and anions do not occur together in solution.
Weak acids acids do not dissociate in solution hence, we can still spot the cations connected to their anions in solution. Hence, the acid having the yellow anion is a weak acid.
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The question is incomplete, complete question is :
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?
Structure is given in an image?
Answer:
There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Explanation:
Total numbers of carbon = 10
Number of primary carbons that is carbon joined to just single carbon atom = 6
Number of secondary carbons that is carbon joined to two carbon atoms = 1
Number of tertiary carbons that is carbon joined to three carbon atoms = 2
Number of quartenary carbons that is carbon joined to four carbon atoms = 1
So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Answer:
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Explanation:
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