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4 years ago

What is the main difference between the Schrödinger model and the Bohr atomic model?

Chemistry

2 answers:

dexar [7]4 years ago

9 0

Answer:

Bohr believed electron orbits were circular, but Schrödinger’s equations represent complex shapes describing the probable locations of electrons.

Explanation:

Nuetrik [128]4 years ago

3 0

The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all locations in space, some being much more likely than others

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Which ONE of the following is an oxidation–reduction reaction? A) PbCO3(s) + 2 HNO3(aq) ––––> Pb(NO3)2(aq) + CO2(g) + H2O(l)

sveta [45]

Answer:

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Explanation:

Which ONE of the following is an oxidation–reduction reaction?

A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.

B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.

C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.

D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.

8 0

4 years ago

Which of these are ionic compounds? Check all that apply.

SIZIF [17.4K]

I think the correct answers from the choices listed above are the first, third and the last option. Ionic compounds are compounds that dissociates into ions when in aqueous solution. From the list, NH4Cl, KF and MgO are the ionic compounds. Hope this answers the question.

5 0

3 years ago

Read 2 more answers

a prescriber has ordered childrens motrin 400 mg po q6h for a child who weighs 60 kg. how many mg/kg is the child receiving

DedPeter [7]

Answer:

6.67 mg/kg per dose ( 26.67 mg/kg per day)

Explanation:

400 mg / 60 kg = 6 2/3 mg/kg per dose

per <em>DAY</em> is four times this number

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2 years ago

Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane

dolphi86 [110]

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as : methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

#SPJ4

8 0

2 years ago

What conditions make G always positive?

Bess [88]

The answer is c ......

8 0

3 years ago

Read 2 more answers