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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original model stated that electrons were tiny little objects circling the nucleus along fixed orbits, however this is no longer valid as we now know that rather "circling" the nucleus at confined orbits, electrons can seem to be everywhere at once.
You might be interested in
(i)What happens when an aqueous solution of sodium sulphate reacts with an aqueous
MAXImum [283]

Explanation:

2nacl+baso4

the orginal equation is

Na2so4+Bacl2

then na react with cl and ba react with so4

4 0
3 years ago
The molar mass of barium nitrate (ba(no3)2) is 261.35 g/mol. what is the mass of 5.30 × 1022 formula units of ba(no3)2? 0.0900 g
liberstina [14]

Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.

<h3>What is mass?</h3>

The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.

Given,

Molar mass of barium nitrate = 261.35 g/mol

If, 6.022 \times 10^{23} have a mass of 261.35 g/mol then, 5.30 \times 10^{22} formula units will have a mass of,

\begin{aligned}& = \dfrac{261.35 \times 5.30 \times 10^{22}}{6.022\times 10^{23}}\\\\&= 23.0\;\rm gm\end{aligned}

Therefore, option C. 23.0 gm is the mass of barium nitrate.

Learn more about mass here:

brainly.com/question/24958554

6 0
2 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
Which atomic model proposed that elections move in specific orbits around the nucleus of an atom? A) Dalton's atomic model B) Th
wel
<span>d) Bohr’s atomic model</span>
6 0
3 years ago
Copper has two naturally occurring isotopes with atomic masses of 62.9296 u () and 64.9278 u (). The atomic mass of copper is 63
natta225 [31]

Answer:

The abundance of first isotope is 69.15 %

The abundance of second isotope is 30.85 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope:

% = x %

Mass = 62.9296 u

For second isotope:

% = 100  - x  

Mass = 64.9278 u

Given, Average Mass = 63.546 u

Thus,  

63.546=\frac{x}{100}\times {62.9296}+\frac{100-x}{100}\times {64.9278}

Solving for x, we get that:

x = 69.15 %

<u>The abundance of first isotope is 69.15 %</u>

<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>

4 0
4 years ago
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