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aev [14]
3 years ago
9

How many moles are in 11.0 grams of methane (CH4)?

Chemistry
1 answer:
Bas_tet [7]3 years ago
5 0
If I’m not mistaken it should be 16.04246.
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The three end stages of stars are Red Giant Phase, Second Red Giant Phase and White Dwarf Phase correct?
solong [7]

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.

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3 years ago
Please,help me in question number 2 and 4.<br><img src="https://tex.z-dn.net/?f=2and4%20%5C%5C%20" id="TexFormula1" title="2and4
svetoff [14.1K]
I think it’s 890 I really don’t know
8 0
3 years ago
How did scientists find out about atoms before technology?
sasho [114]

Answer:

There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.

Explanation:

Hope this helps

6 0
3 years ago
Which one is an accurate description of an erosion?
Sergeeva-Olga [200]

the process of eroding or being eroded by wind, water, or other natural agents

6 0
3 years ago
Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
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