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777dan777 [17]
3 years ago
11

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

the universe would have been different if there had been no dark matter. In your answer, use the term structures.
Please answer
Chemistry
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together.  After the big bang,<em> the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently</em>.

The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.

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What was Thompson’s model of the atom called
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Cyclopropene decomposes to propene when heated to 500 C, calculate rate constant for the first order reaction 0 min 1.48 mmol/L
Dmitry_Shevchenko [17]

Answer:

k = 0.0306 min-1

Explanation:

The table is given as;

Time, Concentration

0 1.48

5 1.27

10 0.98

15 0.84

The integrated rate law for a first order reaction is given as;

ln [A] = -kt + ln [Ao]

where;

[A] = Final Concentration

[Ao] = Initial Concentration

k = rate constant

t = time

In the table, taking the first two sets of values;

t = 5

k = ?

[Ao]  = 1.48

[A] = 1.27

Inserting into the equation;

ln(1.27) = - k (5) + ln(1.48)

ln(1.27)  - ln(1.48) = -5k

-0.1530 = -5k

k = -0.1530 / -5

k = 0.0306 min-1

6 0
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On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical
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Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

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ExplanationBecause:

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