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Makovka662 [10]
3 years ago
10

Two representations of the distribution of electrons in the potassium atom are shown. Model 1 Model 2 1s22s22p63s23p64s1 An oran

ge circle labeled upper K is surrounded by four concentric circles. The innermost circle has dots at the north and south positions. The second and third circles have eight dots, equally spaced, with dots aligning with those on the inner ring. The outer ring has one dot at the north position. Which model is useful in showing the reactivity of potassium?
Chemistry
2 answers:
KengaRu [80]3 years ago
7 0

Answer:

The answer is: both models because each shows an electron in the outermost shell.

Explanation:

vaieri [72.5K]3 years ago
6 0

Answer:

The answer is D.

Explanation:

Please give brainliest

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Which best describes the transition from gas to liquid?
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PRACTICE
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Answer:

35.8 u

Explanation:

The atomic mass of Cl is the weighted average of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its percent abundance).

Atomic mass of Cl-35 = 17p + 18n = 17 × 1.007 u + 18 × 1.009 u

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Atomic mass of Cl-37 = 17p + 20n = 17 × 1.007 u + 20 × 1.009 u

= 17.119 u + 20.180 u = 37.30 u

Set up a table for easy calculation.

0.755 × 35.28 u =  26.64  u

0.245 × 37.30 u =    9.138 u

             TOTAL = 35.8     u

Note: The actual atomic mass of Cl is 35.45 u.

The calculated value above is incorrect because

(a) the given isotopic percentages are incorrect and

(b) the protons and neutrons have less mass when they are in the nucleus than when they are free. Thus, the calculated masses of Cl-35 and Cl-37 are too high.

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How many molecules are 4.3 x 10^27 molecules of N2O5<br>​
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Answer:

<h2>7142.86 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4.3 \times  {10}^{27} }{6.02 \times  {10}^{23} }  \\  = 7142.857...

We have the final answer as

<h3>7142.86 moles</h3>

Hope this helps you

6 0
3 years ago
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