The dipole moments of ad are in the order
The dipole moments of ad are in the order
Option 1 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo>=</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo>=</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> <mo><</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">C</mi> <mi mathvariant="normal">l</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> </span></p>
2 Views|Posted 7 months ago
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7 months ago
Correct Option - 3
Detailed Solution:
Kindly consider the following figure
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
Kindly consider the solution
Fact.
Kindly go through the solution
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