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Both compounds are deprotonated at the sp³-hybridized α-carbon, not at oxygen or sulfur. The resulting conjugate base is stabilized by resonance, with the negative charge delocalized between the α-carbon and the heteroatom.

In compound I, the negative charge is delocalized onto oxygen, whereas in compound II, it is delocalized onto sulfur. Because oxygen is more electronegative than sulfur, it stabilizes the negative charge more effectively. Therefore, the conjugate base of compound I is more stable, making compound I more acidic than compound II.

Although sulfur is larger and more polarizable, that factor is more important when comparing negative charges located directly on oxygen and sulfur, such as RO⁻ versus RS⁻. Here, the strength of the carbonyl-like resonance stabilization makes the oxygen-containing compound more acidic.

So, the correct conclusion is:

Compound I is more acidic than compound II because its conjugate base is better stabilized by resonance involving oxygen.

For the attached question, I thought the most acidic molecule would be 2 because Sulfur can stabilize the electrons better than oxygen, since it is larger. However, the correct answer had the first molecule as the most acidic. I was wondering how that was the case.

Kriti_Mock_Test_2_Combined_Report.pdf

Thank you so much Uma mam

Propane, C₃H₈, is a hydrocarbon. In its complete combustion, the second reactant is O₂, and the products are CO₂ and H₂O.
First, write and balance the equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Now check what is given and what is required.
We are given 1.00 g of propane, and we need to calculate the mass of O₂ consumed.
So, here is the plan:
grams of C₃H₈ → moles of C₃H₈ → moles of O₂ → grams of O₂
First, convert grams of propane to moles using its molar mass. Then convert moles of propane to moles of oxygen using the mole ratio from the balanced equation. Finally, convert moles of oxygen to grams using the molar mass of O₂.
The dimensional-analysis setup is:


All the intermediate units cancel, leaving the desired unit, grams of O₂.

3.63,g,O₂



Since 1.00 g has three significant figures, the final answer should also be reported to three significant figures.

I need help with the following:
**EP 3.14** Propane, C₃H₈, is a common fuel used for cooking and home heating. What mass of O₂ is consumed in the combustion of 1.00 g of propane?

a. 5.00 g

b. 0.726 g

c. 2.18 g

d. 3.63 g

Yes, you are right. The answer should be KNO₃. However, I am unable to see the full question, so I cannot confirm it with complete certainty.

Hi Uma mam, why isn't the answer KNO3 instead of NaCl? For NaCl, around 15 g is recovered and in KNO3, 105 g is recovered.

Mkae the RICE table to see the final answer.
Since the question says equal numbers of moles of HCl and O₂  are taken initially, we can simply assume their initial concentrations are equal.

Let the initial concentration of each reactant be 1.
                        R  4HCl(g) +  O2�(g)⇌ 2Cl2�(g)  +  2H2�O(g)
                                       I     1          1       0           0
                                       C     -4x         -x      +2x        +2x
                                       E      1-4x        1-x     2x        2x
From this its clear that II and III are correct.However I can not be predicted as it depends on the value of x.
So this is not something that must always be true.