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Strike441 [17]
3 years ago
15

what aspect of Bohr's original model of election stucture is no longer considered valid in the currently accepted theory of elec

tron structure
Chemistry
2 answers:
Harlamova29_29 [7]3 years ago
7 0
Bohr suggested circular orbits of electrons around the nucleus of hydrogen atom, but researches have shown that motion of electron is not in a single plane., but it takes place in three dimensional space. Actually, the atomic model is not flat
sergejj [24]3 years ago
5 0
Bohr's original model stated that electrons were tiny little objects circling the nucleus along fixed orbits, however this is no longer valid as we now know that rather "circling" the nucleus at confined orbits, electrons can seem to be everywhere at once.
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What kind of product is formed when a ketone reacts with a grignard reagent followed by protonation?
svet-max [94.6K]

c. a tertiary alcohol; when a ketone reacts with a grignard reagent followed by protonation a tertiary alcohol is formed.

More about tertiary alcohol:

No hydrogen atoms are bonded to the functional group's carbon in a tertiary alcohol. Alcohols that have a hydroxyl group bonded to the carbon atom and are linked to three alkyl groups are referred to as tertiary alcohols. These alcohols' structural makeup largely determines their physical characteristics.

This -OH group's existence enables alcohols to create hydrogen bonds with the atoms next to them. Because of this weak connection, alcohols have higher boiling points than their alkane counterparts.

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8 0
2 years ago
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nasty-shy [4]

Answer:

Eventually, these individual laws were combined into a single equation—the ideal gas ... We find that temperature and pressure are linearly related, and if the ... then P and T are directly proportional (again, when volume and moles of gas are ... of the variables, and they are more difficult to use in fitting theoretical equations ...

Explanation:

3 0
3 years ago
Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

  • <u>Option a:</u>  50.0 g Fe, C_{Fe}=0.449J/g^oC

We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

Change in temperature = 0.96°C

  • <u>Option b:</u>  25.0 g Pb, C_{Pb}=0.128J/g^oC

We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC

Change in temperature = 62.5°C

  • <u>Option d:</u>  25.0 g Ag, C_{Ag}=0.235J/g^oC

We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

Change in temperature = 34.04°C

  • <u>Option e:</u>  25.0 g granite, C_{granite}=0.79J/g^oC

We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0
3 years ago
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