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Tanya [424]
3 years ago
15

Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other

words, do the bond angles change when you switch between Real and Model mode at the top of the page?
Chemistry
1 answer:
blagie [28]3 years ago
5 0

The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6

Answer:

Compounds whose real bond angle are the same as ideal bond angle;

SF6, BF3, CH4, PCI5

Compounds whose real bond angles differ from ideal bond angles;

H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5

Explanation:

According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.

The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.

For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.

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Draw the structural formula of the major product of the reaction of (S)-2,2,3-trimethyloxirane with MeOH, H . Use the wedge/hash
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Answer:

(S)-3-methoxy-3-methylbutan-2-ol

Explanation:

In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (CH_3OH) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.

See figure 1

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8 0
3 years ago
Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?
34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0
3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur
Brums [2.3K]

Answer:

T_2= 31.9\°C

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

-Q_{rxn}=Q_{water}

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

-(-112000J)=m_{water}C_{water}(T_2-T_1)

In such a way, it is possible to define the final temperature as shown below:

T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C

Best regards!

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It makes it more accessible to everybody
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