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Elodia [21]
3 years ago
13

Determine the mass (in grams) of 8.02 x 1024 atoms of mercury (Hg).

Chemistry
1 answer:
Virty [35]3 years ago
3 0
<h3>Answer:</h3>

2670 g Hg

<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>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation: </h3>

<u>Step 1: Define</u>

8.02 × 10²⁴ atoms Hg

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

Avogadro's Number

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 8.02 \cdot 10^{24} \ atoms \ Hg(\frac{1 \ mol \ Hg}{6.022 \cdot 10^{23} \ atoms \ Hg})(\frac{200.59 \ g \ Hg}{1 \ mol \ Hg})
  2. Divide/Multiply:                                                                                               \displaystyle 2671.42 \ g \ Hg

<u>Step 4: Check</u>

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

2671.42 g Hg ≈ 2670 g Hg

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Answer:

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U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

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7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

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