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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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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
3 years ago
If you produce 7 mols of KBr, how many mols of MgBr2 did you need?
djyliett [7]

Answer:

hmm

Explanation:

hmm

7 0
2 years ago
In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made
LenKa [72]
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
8 0
3 years ago
What is four thing the atmosphere does for us
artcher [175]
Keeps in heat, protects us from radiation, provides oxygen, and protects us from objects coming towards the earth.
4 0
3 years ago
Read 2 more answers
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
3 years ago
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