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jeka94
3 years ago
12

How do the energy and the most probable location of an electron in the third shell of an atom compare to the energy and the most

probable location of an electron in the first shell of the same atom?
(1) In the third shell, an electron has more energy and is closer to the nucleus.
(2) In the third shell, an electron has more energy and is farther from the nucleus.
(3) In the third shell, an electron has less energy and is closer to the nucleus.
(4) In the third shell, an electron has less energy and is farther from the nucleus.
Chemistry
1 answer:
solmaris [256]3 years ago
7 0
The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell. The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
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Dimas [21]

Answer:

The final volume of the balloon is = 28.11 L

Explanation:

Initial pressure P_{1} = 1.03 atm = 104.325 K pa

Initial temperature T_{1} = 26 °c = 299 K

Initial volume V_{1} = 22.4 L

Final temperature T_{2} = 22 °c = 295 K  

Final pressure P_{2} = 0.81 atm = 82 K pa

We know that

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Put all the values in above formula we get

\frac{(104.325)(22.4)}{299} = \frac{(82)(V_{2} )}{295}

V_{2} = 28.11 L

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8 0
3 years ago
2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)
Paha777 [63]
<h3>Answer:</h3>

16.7 g H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                               \displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})
  2. Multiply/Divide:                 \displaystyle 16.6685 \ g \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

16.6685 g H₂O ≈ 16.7 g H₂O

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I am not to sure but is it all 4 of them?
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