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IrinaVladis [17]
3 years ago
8

Consider the molecular structure for linuron, an herbicide, provided in the questions below. How many pi bonds are in the molecu

le
Chemistry
1 answer:
Sophie [7]3 years ago
6 0

In the herbicide linuron whose structure is shown in the image attached, there are four pi bonds.

A pi bond is formed by a sideways overlap of atomic orbitals. A sigma bond is formed by an end to end or head to head overlap of atomic orbitals. Pi bonds lead to the occurrence of multiple bonds in the molecule.

In the herbicide linuron whose structure is shown in the image attached to this answer, there are four pi bonds which are easily spotted as double bonds in the structure.

Learn more: brainly.com/question/7291190

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NO + H2-&gt; N2 + H2O<br> Balance equation?
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Answer:

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Explanation:

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Enter the net ionic equation for this reaction. Express your answer as a net ionic equation. Identify all of the phases in your
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Answer:

2H+(aq) + 2OH-(aq) → 2H2O(l)

Explanation:

Step 1: The balanced equation

2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)

This equation is balanced, we do not have the change any coefficients.

Step 2: The netionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)

After canceling those spectator ions in both side, look like this:

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Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.
BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

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

Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

<u>Step 3: Use Dimensional Analysis</u>

<u />6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} ) = 511.901 g NO₂

<u>Step 4: Check</u>

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

511.901 g NO₂ ≈ 510 g NO₂

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wolverine [178]
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