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NNADVOKAT [17]
3 years ago
13

Balance the reactions which form ions. Choose "blank" if no other coefficient is needed. Writing the symbol implies "1."

Chemistry
2 answers:
ryzh [129]3 years ago
8 0
The answers are the following:
1) (NH4)2CO3 2 NH4 +→ + CO3 -2
2) PbI2 →1 Pb+2 +2 I+1
3) (NH4)3PO4 →3NH4 + 3 + PO4 +
Minchanka [31]3 years ago
7 0
Following are the Balanced Reactions,

Reaction 1:


                            (NH₄)₂CO₃     →    <u>2</u> NH₄⁺  +  CO3⁻²

Reaction 2:

                                     PbI₂    →    Pb⁺²<span>  +  <u>2</u> I</span>⁻¹

Reaction 3:

                              <span>(NH</span>₂<span>)</span>₃<span>PO</span>₄     →     <span><u>3</u> NH</span>₄⁺  <span>+   PO</span>₄⁻³
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Which one of the following compounds will NOT be soluble in water?
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675 g of carbon tetrabromide is equivalent to how many
VARVARA [1.3K]
<h3>Answer:</h3>

2.04 mol CBr₄

<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>Organic</u>

  • Writing Organic Compounds
  • Writing Covalent Compounds
  • Organic Prefixes

<u>Atomic Structure</u>

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

<u>Step 1: Define</u>

675 g CBr₄

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

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

<u>Step 3: Convert</u>

<u />\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4<u />

<u />

<u>Step 4: Check</u>

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

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

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How many electrons are in a s-2 ion​
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