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g100num [7]
3 years ago
7

What observation led chadwick (and rutherford) to conclude there must be something besides just the proton in the nucleus of ato

ms?
Chemistry
2 answers:
ludmilkaskok [199]3 years ago
8 0

Rutherford used the gold foil experiment to disprove Thompson Plum Pudding model of the atom. Rutherford experiment showed that the center of the atom contains a positively charged nucleus and most of the atom’s mass lies in the center of the atom in a small volume. Later Chadwick proved that the nucleus of the atom also contains uncharged particles called neutrons.  

adoni [48]3 years ago
5 0
Rutherford discovered the nucleus, and Chadwick discovered the neutrons. Before their discoveries, the widely accepted theory was Thompson's where the electrons and protons are located all inside the nucleus. However, when Rutherford did his gold foil experiment, some of the cathode rays (negatively charged) were reflected back to the source. This is due to the presence of the dense nucleus where the neutron resides.
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what is the relationship between the dipole moment and bond moment? explain it with examples ,how it possible for a molecule to
Mariulka [41]

Answer :

Dipole moments occur when there is a separation of charge. Dipole moments occur due to atoms electronegativity, where one atom has the ability to attract electrons towards it giving it a negative charge and the one deficient in electrons acquire a positive charge called as the bond moment.

But if the bond moments are equal and opposite in direction , they cancel each other and thus there is no net dipole moment in the molecule.

For example: In carbon dioxide , both the -C=O bonds are polar but as the molecule is linear and the the magnetic moments are equal and oppposite, they cancel each other and the molecule is non polar.

4 0
3 years ago
Determine the expression for the equilibrium constant, kc, for the reaction by identifying which terms will be in the numerator
MakcuM [25]

The kc is a representation of how fast the reaction proceeds to their products when it has achieved equilibrium. The activation energy for the forward and the one for the reverse reaction are similar because they attained chemical equilibrium. A chemical equilibrium happens when both of the reactant and products achieve the same concentration. An example is the process of melting and freezing. Melting and freezing for a given substance occurs at the same temperature. Because the temperature at which the solid starts to melt is also the temperature at which the liquid starts to freeze. They are at chemical equilibrium.

3 0
4 years ago
Fluorine is a gas and iodine is a solid, yet all halogens are non-polar and would interact with London dispersion forces. This i
attashe74 [19]

Answer:

the molecules of each substance attract each other through dispersion (London) intermolecular force.whether a substance is solid, liquid or gas depends on the balance between the kinetic energies of the molecules and their intermolecular attractions.

thank you.

6 0
3 years ago
Calculate: (7.0 x 10^5) + (2.5 x 10^4) =
Serjik [45]

Answer:

725000

Explanation:

hope you find this helpfull

8 0
3 years ago
Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net
Degger [83]

Answer:

- AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)

- K=1.2x10^{-5}

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}

And the formation of the complex:

Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7

We obtain the balanced net ionic equation by adding the aforementioned equations:

AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}

Best regards.

4 0
3 years ago
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