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baherus [9]
3 years ago
13

Why doesn't the universe collapse under the force of gravity? Please tell where you got the information from and if it is correc

t, you'll get points! :3
Chemistry
2 answers:
Sergio [31]3 years ago
8 0
First, this would go under physics. 

Gravity is a relatively weak force and depends on the mass and distance of two objects. Even between insects, the force of gravity exists. However, it is not strong enough to be present because the two masses are too small and too far away.

Here is a Harvard question that is similar.
https://www.cfa.harvard.edu/seuforum/faq.htm
If the universe started out so dense, why didn't it collapse into a black hole?
A large enough clump of matter will collapse to form a black hole, but ONLY if it is surrounded by (relatively) empty space. During the Big Bang, there WAS NO empty space: ALL of space was filled more or less uniformly with matter and energy; there was no "center of attraction" around which matter could coalesce. Under these circumstances, a cosmic-scale black hole will not form (and lucky for us!).
liberstina [14]3 years ago
7 0
According to Newton's 3rd law if body a exerts a force them body b exerts a force that is equal but opposite. Therefore the universe exerts a force on everything that is inside the universe and the moon,planet, stars etc. exerts a force back on the universe.

According to http://www.kickassfacts.com/askus-after-the-big-bang-why-didnt-the-universe-re-collapse-under-its-own-self-gravity/


"It certainly slowed down under its own gravity, but not enough to recollapse.

There’s a very simple (and almost exact) analogy. Let’s represent the Big Bang by launching a rocket. For our purposes, it isn’t propelled at all after the moment of launch, but of course, initially it’s shot up at some very high speed. Your question is exactly analogous to asking why the rocket didn’t fall back down to the Earth.

The answer is that the rocket was launched with an initial speed greater than or equal to the Earth’s escape velocity. As the rocket moves up, gravity does slow it down, but gravity also gets weaker. Escape velocity is the speed where gravity weakens more quickly than it can slow the rocket to a halt. So even though the Earth’s gravity is certainly slowing it down as it goes up, it never slows it down so much that it stops and falls back down to the ground.

We can map this answer directly onto the expanding Universe. Why hasn’t it recollapsed under its own gravity? Because even though the gravity of all the matter and energy in the Universe does cause the expansion to slow down (or at least did, until recently), it was initially expanding so quickly that, like a rocket moving up at escape velocity, it never slowed down quite enough to stop and recollapse."

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If the atomic number of an element is 9 and the mass number is 19, how many neutrons does the atom have? 10 9 19 28 1 points
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6 0
4 years ago
50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the
scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2  +  H^+  ------>  H_3}^+NC_2H_5CO^-_2

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = \frac{initial moles of alaninate}{total volume}

[H_3}^+NC_2H_5CO^-_2] = \frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}

[H_3}^+NC_2H_5CO^-_2] = \frac{8}{100mL}

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = \sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{  K_{a1}{[H_3}^+NC_2H_5CO^-_2]  } }

[H^+] = \sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})}  {(10^{-pK_{a1}})+(0.08)} }

[H^+] = \sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})}  {(10^{-2.344})+(0.08)} }

[H^+] =  7.63*10^{-7}M

pH = - log [H^+]

pH = -log[7.63*10^{-7}]

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = \frac{initial moles of alaninate}{total volume}

[H^+_3NC_2H_5CO_2H] = \frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}

[H^+_3NC_2H_5CO_2H] = \frac{8}{150}

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H        ⇄        H^+_3NC_2H_5CO^-_2    +  H^+

(0.053 - x)                                  x                             x

K_{a1} = \frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}

10^{-PK_{a1}} = \frac{x*x}{(0.053-x)}

10^{-2.344} =\frac{x^2}{(0.053-x)}

0.00453 = \frac{x^2}{(0.053-x)}

0.00453(0.053-x) =x^2

x^2+0.00453x-(2.4009*10^{-4})

Using quadratic equation formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

we have:

\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)} OR \frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0
3 years ago
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