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3 years ago

Why do an atoms electrons revolve around its proton instead of drifting away?

Chemistry

1 answer:

WINSTONCH [101]3 years ago

8 0

Answer:

Electrons are pulled to the nucleus by gravity. The modern atomic model describes the electrons as moving around the nucleus like a cloud, rather than in perfect orbits. I think atoms and protons attract but i'm not 100% sure on that haha. I hope this helps you out :)

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Buckets of paint can be classified as what type of mixture?

marysya [2.9K]

Answer:

Homogeneous

Explanation:

Homogeneous mixtures are uniform in composition. They have the same proportion of components throughout. Homogeneous mixtures are called solutions. Sugar, paint, alcohol, gold are all examples of homogeneous mixtures because they look the same throughout.

6 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

In general, the more _ a metal has, the stronger its metallic bonds will be because...

Nuetrik [128]

In general, The more valence electrons a metal has, the stronger its metallic bonds will be because Boron is a metalloid and is ionically bonded.it is too electronegative to release its valence electrons for metallic bonding.As a result, their valence electrons feel a stronger pull from the nucleus (a greater effective nuclear charge) and are less easily released for metallic bonding.

5 0

3 years ago

If you drink at 2am can you pass a drug test at 6pm

VikaD [51]

Yeah...it should be out of your system by then

3 0

3 years ago

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Calculation of Original pH from Final pH after Titration A biochemist has 100 mL of a 0.10 M solution of a weak acid with a pKa

ddd [48]

Answer:

Explanation: