When NaCl dissolves in water

1 answer:

6 0

The sodium ion becomes hydrated. when sodium chloride dissolves in water, the sodium and chloride ions and the polar water molecules are strongly attracted to one another by ion-dipole interactions

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A 50.0-gram block of copper at 10.0°C is carefully lowered into 100.0 grams of water at 90.0°C in an insulated container. Which

Andrei [34K]

First consider and take note of the following:
1. When 2 or more finite amounts of substances are mix together (with different temperature) the final temperature is never equivalent to the initial temperature of any of the substances involved.
2. The substances meet at an equilibrium temperature.
3. Heat transfers from a higher temperature to a lower temperature.

Only choice (3) satisfies the conditions.

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3 years ago

How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?

kozerog [31]

its d because it all has to be balanced

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Write electronic configuration of chlorine in its ionic state?

MrMuchimi

Answer:

1s2 2s2 2p6 3s2 3p6

Explanation:

Chlorine is a groups 17 element. The halogens for ions by accepting one electron to form univalent negative ions.

Since chlorine normally contains seventeen electrons, the chloride ion consists of eighteen electrons.

Hence the electronic configuration of chlorine ion is; 1s2 2s2 2p6 3s2 3p6.

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3 years ago

P-fluoroanisole reacts with sulfur trioxide and sulfuric acid. Draw the major product of this substitution reaction; if applicab

Ipatiy [6.2K]

Answer: (Structure attached).

Explanation:

This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.

The aromatic ring in p-fluoroanisole has two sustituents, an halogen and a methoxy group, which are ortho-para directing substituents.

Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using sulfur trioxide as an electrophile (very reactive).

The reaction occurs in three steps:

The attack on the electrophile forms the sigma complex.
The loss of a proton regenerates an aromatic ring.
The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).

Normally, a mixture of ortho-para substituted products would be obtained. However, since both para positions are occupied, only the ortho substituted product is obtained here.

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Students rolled a glass marble and a baseball down an inclined plane, as shown below, and made observations.

Alja [10]

The answer is b i just took the quiz and got it right|| k12.gca student

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3 years ago

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