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lisabon 2012 [21]
4 years ago
12

How can you which molecule is positive and negative in a dipole-dipole relationship?

Chemistry
1 answer:
marysya [2.9K]4 years ago
5 0
Dipole interactions are observed in covalent bonds. In ionic bonding, permanent transfer of electrons occurs and due to this dipole-dipole interactions are not observed. In covalent bonding, electron cloud is shared between 2 atoms. If this electron cloud is not shared equally between them, polarities are formed in a molecule. And hence we say that the molecule is polar. For a molecule to be polar, there should be electronegativity difference between them. Atom with greater electronegative attracts electron cloud more towards itself whereas atom with lesser electronegative attracts electron cloud less. But there is no permanent transfer of electrons. Due to this electronegativity differences, atom with more electronegative gains partial negative charge and atom with lesser electronegative value gains partial positive charge. The charge is partial because there is no complete transfer of electrons.
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Below is a short reaction sequence. Provide the structure of the two missing organic compounds.
dimaraw [331]

The product of this reaction is a halohydrin as shown here.

<h3>What are the products?</h3>

We have a reaction that first involves the formation an alkene as the bases are used on the first substrate. The alkene that is thus created is now able to react with the bromine in water.

The first step of the reaction is where the multiple bond is created and this multiple bond is what can now go on to participate in a chemical reaction in the next step of the process of reaction as shown in the image.

Bromine in water is also hat we call bromine water. This bromine water is able to add across a double bond and when that happen we will have a saturated compound. This could also be regarded as halohydrin reaction.

Recall that the first step of the reaction of the alkene with bromine is the formation of a dibromide via the brominium intermediate. This is now followed by reaction with water to form the halohydrin product.

Learn more about organic reaction:brainly.com/question/9585105

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4 0
2 years ago
Which change of state is shown in the model?
Romashka [77]
I believe the change of state shown in the model is deposition. 
Deposition is a process in which gases change phase and turns directly in solids without passing through the liquid phase. It is the opposite of sublimation.
One of the major difference between gases and solids is the distance between molecules; in gases the inter molecular spaces are large, while in solid they are very small, making solids be the most dense, with closely packed molecules. This is evident in the diagram, the phase changed from gases to solids. 
5 0
3 years ago
Read 2 more answers
Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi
sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Entropy change of the reaction ; ΔS

\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]

=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0
3 years ago
How many pints are in a 48 cups?
Feliz [49]
1 cup = 2 pints

48 × 2 = 96 pints

HOPE THIS HELPED
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3 years ago
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kifflom [539]
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