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spayn [35]
3 years ago
6

How many electrons can be contained in the second energy level?

Chemistry
2 answers:
slamgirl [31]3 years ago
6 0
I thought it was 4 for the second layer, then 8, 12, etc
Lelechka [254]3 years ago
3 0
Its D for the second energy level, but the first energy level is 2.
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<span>Solids, Liquids, Gases, Plasma, and Bose-Einstein Condensates. The main differences between these states of matter are the densities of the particles.</span>
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3 years ago
What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?
astraxan [27]

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.  

The law that we can applied to calculate the M is:  

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;  

M = 0.082 mol  /  2.03 L = 0.04 mol/L

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3 years ago
If a “sample” of pennies contained 75 heads and 25 tails, how many half‐lives would have passed since the “sample” formed. Expla
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8 0
2 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}

=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

3 0
3 years ago
Is this correct please help?
iren [92.7K]

Answer:

\boxed {\tt Number \ and \ unit}

Explanation:

The two requirements for a measurement are a <u>number</u> and a <u>unit.</u>

For example, here is a measurement:

38.6 cm

The <u>number</u> is 38.6 and the <u>unit</u> is cm, or centimeters.

Therefore, both <em>number </em>and <em>unit</em> are correct.

7 0
3 years ago
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