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Andrew [12]
3 years ago
11

Let’s start by finding the location on the periodic table for the atoms for substances that conduct when dissolved. What do you

notice about where the atoms are?
Chemistry
1 answer:
levacccp [35]3 years ago
3 0

Answer:

Explanation:

This question appear incomplete. However, chemical substances that conduct electricity when dissolved in water are elctrovalent/ionic compounds. These compounds are usually made up of elements in group 1 and group 7 (for example NaCl, NaI and LiF) but could also be made up of group 2, group 3 and group 6 [for example Mg(OH)₂ and Al(OH)₃]. Thus, they can be loosely said to be found in those groups in the S and P blocks of the periodic table.

NOTE:

Sodium (Na) is found in group 1 (S-block)

Lithium (Li) is found in group 1 (S-block)

Chlorine (Cl) is found in group 7 (P-block)

Iodine (I) is found in group 7 (P-block)

Flourine (F) is found in group 7 (P-block)

Magnesium (Mg) is found in group 2 (S-block)

Aluminium (Al) is found in group 3 (S-block)

Oxygen (O) is found in group 6 (P-block)

Hydrogen (H) is found in group 1 (S-block)

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7 0
3 years ago
2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2
miv72 [106K]
<h3>Answer:</h3>

4 g AgCl

<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
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN]   2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given]   5.0 g AgNO₃

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

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})
  2. Multiply/Divide:                                                                                                  \displaystyle 4.21533 \ g \ AgCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

4.21533 g AgCl ≈ 4 g AgCl

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