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nikitadnepr [17]
3 years ago
6

Bohr found experimental evidence for his atomic model by studying _____.

Physics
2 answers:
nevsk [136]3 years ago
6 0
For the answer to te question above, it is a. The Electron Cloud.
For additional information, The Electron Cloud is an informal term in physics. It is used to describe where electrons are when they go around the nucleus of an atom.
I hope my answer helped you.
frozen [14]3 years ago
5 0

Answer:

b)  line spectra

Explanation:

 The model of the atom as given by Rutherford was modified by Bohr. According to Rutherford's model, the electrons that orbit the nucleus electromagnetic radiation. As a result they lose energy and had to collapse into the nucleus, thereby making the atom unstable.

Bohr corrected that by proving that electrons revolve around the nucleus only in certain fixed orbits, with a specific energy. They can jump from one to another and lose or gain energy by emitting or absorbing energy in that process.

Electron cloud is the region where the electrons are most likely to occur. This was proposed by Schrodinger.

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What does the model below represent?
serious [3.7K]

Answer:

B. a chemical reaction

Explanation:

2CO + O2 → 2C02

8 0
3 years ago
Read 2 more answers
• List four uses of the concave mirror.<br> State five uses of lenses in everyday activities
Zina [86]
FOUR USES OF CONCAVE MIRROR:Satellite dishes,headlights of a car, telescopes used for astronomical studies, and shaving mirrors because of there curved and reflective surface.

FIVE USES OF LENSES: Camera lens ,microscopes ,magnifying glass,eyeglasses,projector
6 0
2 years ago
An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c
Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

           Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

8 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo
vovikov84 [41]

Answer:

L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

So,

we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δv^{2}  

52000 J = (0.5) x (C) x (610^{2})

C = 0.28 F

And we also know that,

C = \frac{K*E*A}{d}

E = 8.85 x 10 ^{-12}

K = 3.7

A = 0.20 x L

d = 2.6 x 10^{5} m

Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

Solving for L, we get:

L = 1.11 x 10^{6} m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

7 0
3 years ago
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