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Sindrei [870]
3 years ago
14

The three end stages of stars are Red Giant Phase, Second Red Giant Phase and White Dwarf Phase correct?

Chemistry
1 answer:
solong [7]3 years ago
6 0

Answer: Most of the stars in the universe are main sequence stars — those converting hydrogen into helium via nuclear fusion. A main sequence star may have a mass between a third to eight times that of the sun and eventually burn through the hydrogen in its core. Over its life, the outward pressure of fusion has balanced against the inward pressure of gravity. Once the fusion stops, gravity takes the lead and compresses the star smaller and tighter.

Temperatures increase with the contraction, eventually reaching levels where helium is able to fuse into carbon. Depending on the mass of the star, the helium burning might be gradual or might begin with an explosive flash.

You might be interested in
How many gallon are there in 85.00 L?
DerKrebs [107]

Answer:

22.46

Explanation:

.There are 3.79 liters in one gallon

4 0
2 years ago
A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra
Rasek [7]

The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

First-order reactions are very commonplace. we have already encountered  examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁  = 0. 543 /s

T₁  = 25°C

Activation energy Eₐ =  75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁=  Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation  

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.

Learn more about activation energy here:- brainly.com/question/26724488

#SPJ4

4 0
1 year ago
Given 7.45 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100
KengaRu [80]

The equation for the reaction is:

C₄H₈O₂ + C₂H₅OH = C₆H₁₂O₂ + H₂O

Now you see that the number of the moles of butanoic acid and etyl butyrate is equal in

the reaction. That means;

number of moles of C₄H₈O₂ = number of moles of C₆H₁₂O₂

mass of C₄H₈O₂/ Molar mass of C₄H₈O₂ = mass of C₆H₁₂O₂/ molar mass of C₆H₁₂O₂

mass of C₆H₁₂O₂ = molar mass of C₆H₁₂O₂ x mass of C₄H₈O₂/ Molar mass of C₄H₈O₂

Now, assuming <span>100% yield, the mass of ethyl butyrate produced is: </span>

<span>= 7.45/88.11 x 116.16</span>

<span>=9.82g</span>

<span>Thus, the theoretical yield of ethyl butyrate is 9.82g.</span>

3 0
2 years ago
You are going to standardize your sodium hydroxide by titrating with potassium hydrogen phthalate. As an example, you dissolve 0
Kay [80]

Answer:

0.13 M

Explanation:

The reaction equation is;

NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)

Molar mass of KHP = 204.22 g/mol

Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles

n= CV

Where;

C= concentration

V= volume in dm^3

n= number of moles

C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M

If 1 mole of KHP reacts with 1 mole of NaOH

1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH

From

n= CV

We have that only 12.44 ml of NaOH reacted

C= n/V = 1.65 × 10^-3 moles × 1000/12.44

C= 0.13 M

At the equivalence point, the KHP solution turned light pink.

4 0
3 years ago
What happens if ..increasing the atomic number in period 1 (concerning the atomic size)​ .
polet [3.4K]

When the atomic number increases The atomic size will be larger.

7 0
2 years ago
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