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DochEvi [55]
3 years ago
8

A student places a small amount of water in a glass dish and then carefully places a small paperclip on the surface of the water

. The paperclip floats. What happens if the student adds a drop or two of dish soap to the water? Explain your answer.
Chemistry
2 answers:
Ipatiy [6.2K]3 years ago
6 0

I believe that adding soap will make the paperclip fall through the water to the base of the dish. Soap is a surfactant, which decreases the surface pressure of the water. The surface strain of water is what upheld the paperclip in the first place, so with the decrease in surface pressure the paper clip will fall.

rosijanka [135]3 years ago
3 0
Include: 
- Adding cleanser makes the paperclip fall through the water to the base of the dish. 
- Soap is a surfactant. 
- Surfactants lessen the surface pressure of a fluid. 
- The surface strain of water is the thing that upheld the paper cut.
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Arrange the following molecules in order of decreasing polarity of their bonds.
polet [3.4K]

Answer:

SF2 > H2O > PBr3 > NCl3

Explanation:

Compare the electronegativity values for the atoms and classify the nature of the bonding based on the electronegativity difference.

P has an electronegativity of 2.1, while Br has an electronegativity of 2.96. The difference is 0.86, indicating that these atoms will form covalent bonds.

S has an electronegativity of 2.58 while F has an electronegativity of 4.0. The difference is 1.42, indicating that these atoms will form polar covalent bonds.

O has an electronegativity of 3.5 while H has an electronegativity of 2.1. The difference is 1.4, indicating that these atoms will form polar covalent bonds.

N has an electronegativity of 3.04, whereas Cl has an electronegativity of 3.5. This difference of 0.46 indicates that these atoms will form covalent bonds.

We know that the greater the electronegativity, the higher the polarity. In decreasing order of polarity, we have:

SF2 > H2O > PBr3 > NCl3

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The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.

The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).

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The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.

Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.    


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