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tiny-mole [99]
3 years ago
9

Fluorine (F) has seven electrons in its outermost shell. Fluorine _____.

Chemistry
2 answers:
Ira Lisetskai [31]3 years ago
6 0
2. F is non-metal, because it has seven electrons (more that 3)
3.It <span>fills its shell by gaining one electron, because it is easier to gain 1 e⁻ than to loose 7 e⁻.
4. It </span><span>becomes a negatively charged ion, because atom of F is neutral, it has "0" charge, and it takes one negatively charged electron(-1). (0 +(-1)= -1)</span>
Anika [276]3 years ago
3 0

Answer:

is a nonmetal

fills its shell by gaining one electron

becomes a negatively charged ion

Explanation:

Fluorine is in group 7 of the periodic table and in period 2. Elements found in group 7 are mostly non metals. This non metals mostly gain electron during bonding with other atom of elements.

Fluorine fills it outer shell by gaining a single electron. The valency electron of Fluorine is 7 . It can easily attract 1 electron to it self to attain the octet rule. This is why most non metal in group 7 are very reactive .

Fluorine easily become an anion as they gain single electron. The charge become negatively charged. It gains single charge and possess a charge of -1.

The electronic configuration is 2 7. The atomic number of Florine is 9.

 

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In a 3.7- gram sample, there are______ atoms of gold
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<h3>Answer:</h3>

1.1 × 10²² atoms Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<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
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.7 g Au

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

Avogadro's Number

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

<u />3.7 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au} )(\frac{6.022 \cdot 10^{23} \ atoms \ Au}{1 \ mol \ Au} ) = 1.13121 × 10²² atoms Au

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au

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