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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original model stated that electrons were tiny little objects circling the nucleus along fixed orbits, however this is no longer valid as we now know that rather "circling" the nucleus at confined orbits, electrons can seem to be everywhere at once.
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The concentration of reactant doesn't affect the activity
pashok25 [27]

Answer:

B.false because if the reactant concentration is disturbed the whole reaction will be affected.

Explanation:

6 0
3 years ago
Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
A small hole in the wing of a space shuttle requires a 20.7-cm² patch.
cricket20 [7]

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or,  1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 ×  10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

#SPJ4

8 0
2 years ago
(a) Show that the pressure exerted by a fluid P (in pascals) is given by P = hdg, where h is the column of the fluid in metres,
Komok [63]

Answer:

B) THE DEPTH OF THE LAKE IS 0.060 m

Explanation:

b) Determine the depth of the lake in metres

1. Using the general gas law, we will calculate the initial pressure of the air bubbles.

P1V1 /T1 = P2V2/T2

P1 = Unknown

T1 = 5.24 °C

T2 = 18.73 °C

P2 = 0.973 atm

V1 = V1

V2 = 6V1

P1 = P2 V2 T1 / V1 T2

P1 = 0.973 * 6V1 * 5.24 / V1 * 18.73

P1 = 5.09852 * 6 / 18.73

P1 = 30.59112 / 18.73

P1 = 1.633 atm.

2. Calculate the depth of the lake:

             Pressure = length * density * acceleration

            length = Pressure / density * acceleration

Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2

Density =  1.02 g/cm3 = 1.02 * 10^3 kg/m^3

Acceleration = 9.8 m/s^2

So therefore, the length in metres is:

Length = density * acceleration / pressure

Length =  1.02 *10^3 * 9.8 / 165, 463.725  

Length = 9.996 * 10^3 / 165 463.725

Length = 0.06 m

Hence, the depth of the lake is 0.06 m

6 0
3 years ago
What two general products do combustion reactions create? *
Dimas [21]
It would be carbon dioxide and water.

hope this helps you
7 0
3 years ago
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