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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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2 moles of substance A contains 12.04 × 1023 atoms.
professor190 [17]

Answer:

2 moles of Substance B

Explanation:

Because 1 mole of any substance contains 6.02× 10²³ particles either atoms, molecules, ions or even electron(Avogardo Constant)

Therefore 12.04 × 10²³/ 6.02× 10²³  =  2 mole of Substance B

7 0
2 years ago
Journal making expressing qualitative data and quantitative data.
barxatty [35]
Journal making expressing data and quantitative data

6 0
2 years ago
Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.
goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}

In the first solution:

X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625

X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375

340torr = 0.625P^0_{A}+0.375P^0_{B} <em>(1)</em>

For the second equation:

X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700

X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300

370torr = 0.700P^0_{A}+0.300P^0_{B}<em>(2)</em>

Replacing (2) in (1):

340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})

340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}

-122.5torr = -0.250P°A

P^0_{A} = 490 torr

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

370torr = 0.700*490torr+0.300P^0_{B}

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 0
3 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
You find small pieces of ice instead of ice cubes in the freezer. State how this is possible.
Nat2105 [25]
Well, if u had a spilled liquid in there (we'll simply go with water) and you had the freezer at a cold temperature it would change (like,icycles on trees when it's snowing)
4 0
3 years ago
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