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Anika [276]
3 years ago
15

Describe the main difference between the Bohr model and the electron cloud model of the atom

Chemistry
2 answers:
Leno4ka [110]3 years ago
4 0

Answer:

The main difference between the two models is <em>the position of the electron in the atom</em>.

Explanation:

  • <em>Bohr model:</em> The electrons are moved around the nucleus in circular definite paths (orbitals or shells). Also, he could not find or detect the exact position of electron.
  • <em>Electron cloud model:</em> It is supposed by  Erwin Schrodinger. He showed that the emission spectra of the atom is the way to detect the probability of electron position.
lara31 [8.8K]3 years ago
4 0

Answer: The main difference is about the certainty of the location of an electron.

Explanation:

The cloud model describe an energy level as region where the likelihood of finding an electron is high.

While the Bohr model describes an energy level as the movement of electron around nucleus of an atom, using the movement of the planets around the sun as an analogy.

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What is the pOH of 0.50 molar H3BO3?
Crazy boy [7]

<u>Answer:</u>

<em>A. 10.25</em>

<em></em>

<u>Explanation:</u>

Pkb =4.77

So pka = 14 - pka = 9.23

Ka =10^{-pka}

H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

Ka =\frac {((x)(x))}{(0.50M-x)}

5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}

(-x is neglected) so we get

5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}

x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}

pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

4 0
3 years ago
Which agent of erosion can create a limestone cave?
Ivanshal [37]
Water containing carbonic acid and calcium
5 0
3 years ago
A chemist is testing the quality of a wine produced by a vineyard. Which branch of chemistry might the chemist use? physical che
Phantasy [73]

Answer:

organic chemistry

Explanation:

There are many branches of chemistry. Some of them are physical chemistry, analytical chemistry, inorganic chemistry and organic chemistry.

A chemist is testing the quality of a wine produced by a vineyard. It means he will test the structure, properties, and preparation of wine. All these tests comes under organic chemistry.

Hence, the correct option is (d) "organic chemistry"

7 0
4 years ago
Read 2 more answers
Exactly 15.0 g of a substance can be dissolved in 150.0 g of water what is the solubility of the substance in grams per 100 g of
Leokris [45]
<span>(15.0 g) / (150.0 g) x (100 g) = 10.0 g/100 g H2O </span>
5 0
3 years ago
When 240 mg of a certain molecular compound X are dissolved in 35.0 g of dibenzyl ether ((C6H5CH2)2O), the freezing point of the
DedPeter [7]

Answer: MM = 16.55 g/mol

Explanation: <u>Freezing</u> <u>point</u> <u>depression</u> is a phenomena that explains why adding a solute to a solvent decreases the solvent freezing point: when a substance begins to freeze, its molecules slows down and rearrange itself forming a solid. If a solute is added, the molecules from the solvent interfere in the formation of the solid. To guarantee the transformation, the solution has to cooled down even more.

Freezing point and molality concentration is related by

\Delta T=T_{f}_{(solvent)}-T_{f}_{(solution)}=K_{f}.m

where

ΔT is freezing point depression

T_{f}_{(solvent)} and T_{f}_{(solution)} are freezing point of solvent and solution, respectively

K_{f} is freezing point depression constant

m is molality concentration

<u />

Dibenzyl ether is the solvent and has the following properties: K_{f}= 6.27 and T_{f} = 3.6°C.

Molality concentration is

m=\frac{T_{(solvent)}-T_{(solution)}}{K_{f}}

m=\frac{3.6-1}{6.27}

m = 0.415

<u>Molality</u> <u>concentration</u> is moles (n) of solute dissolved in a mass, in kilogram, of solvent.

m=\frac{moles}{mass(kg)}

n = m(mass of solvent in kg)

n = 0.415(0.035)

n = 0.0145

<u>Molar</u> <u>mass</u> (M) is the weight of one sample mole and can be calculated as

n=\frac{m}{M}

M = \frac{m}{n}

m in grams

Molar mass of compound X is

M=\frac{0.24}{0.0145}

M = 16.55

<u>Molar mass of molecular</u><u> compound X is 16.55g/mol</u>

3 0
3 years ago
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