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Alex787 [66]
1 year ago
9

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 if you do i will mark you as brainliest
Chemistry
1 answer:
Alecsey [184]1 year ago
3 0

Answer:

Without dark matter galaxies would loose an extreme amount of gas required to create stars.

Without dark matter the universe wont have as many galaxies clumped together forming larger versions of those galaxies. This would cause a change in the structure of the "skeleton" of the web.

(Hope this can help, I didn't do exactly as it is said to because that is your job)

:)

Explanation:

Forbes gives somewhat of an explanation if you are curious.

(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)

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Oxides of nitrogen are pollutant gases which are emitted from car exhausts.
V125BC [204]
Let's look at the molar weight of the answers: 
NO is 30 g/mol 
NO2 is 46 
N2O is 44 
N2O4 is 124 

<span>We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. </span>
You were given the liters (.120L) 
Std pressure is 1 atmosphere 
You&#x27;re looking for n, the number of moles 
<span>Temp is 293.15 kelvin, thats standard </span>
And r is the gas constant in liters-atm per mol kelvin 

(.120 liters)(1atm)=n(293.15K)(.08206) 
Solving for n is .0049883835 mol 

<span>.23g divided by .0049883 mol is about 46g&#x2F;mol. You&#x27;re answer is B I think, NO2

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
4 0
2 years ago
Read 2 more answers
What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr
SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, CH_3CH_2CH_3 the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= 1s^{2}2s^{2}2p^{2}

In excited state: 1s^{2}2s^{1}2p^{3}

Hydrogen (atomic number=1): 1s^{1}

All the bonds in the compound is single bond(\sigma-bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in 1s^{1} of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is s-hybridization and the hybridization of Carbon is sp^{3}-hybridization.

The hybridization of each atom is shown in the image.

3 0
2 years ago
Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t
densk [106]

Answer:

Close to the calculated endpoint of a titration - <u>Partially open</u>

At the beginning of a titration - <u>Completely open</u>

Filling the buret with titrant - <u>Completely closed</u>

Conditioning the buret with the titrant - <u>Completely closed</u>

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

3 0
2 years ago
20 PTS!!! What is the volume of a gas at 90.0°C, if it occupies 1.41 L at 170°C?
Margarita [4]

Answer:

1.72 L

Explanation:

Initial Temperature of the gas T =90°+273= 363 K

Initial Volume of the gas V1= 1.41 L

Final temperature of the gas T2= 170°c +273= 443 K

Final volume of the gas V2= ????

Using Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 1.41×443/363

V2= 1.72 L

3 0
2 years ago
Read 2 more answers
What is the pressure of a 3.00 L gas vessel that has 18.0 grams of helium at 25°C? (R= 0.0821 L atm/ mol K)
nasty-shy [4]
This is an ideal gas law question. You need to use the equation PV=nRt. First you need to find n, the number of mols of helium. The molar mass of helium is 4.00g/mol, and 18g/4.00g/mol = 4.5 mols of helium. Next you need to convert the temperature from Celsius to Kelvin, because only kelvin temperatures can go into the deal gas law equation. To convert 25C to kelvin, add 273. That gives you 298K. Now you can plug all of you information into the ideal gas law equation and solve for P, pressure.
PV=nRt
P(3.00L)=(4.5mol)(0.0821LatmbmolK)(298K)
P=36.70atm
Please give brainliest if this was helpful!
7 0
2 years ago
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