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Alecsey [184]
4 years ago
14

Classify each substance based on the intermolecular forces present in that substance. Hydrogen bonding, dipole–dipole, and dispe

rsionDipole–dipole and dispersion only Dispersion only
H20
CH4
CO
CH3Cl
Chemistry
1 answer:
mamaluj [8]4 years ago
3 0

Let us understand the inter molecular forces in brief

a) hydrogen bond: it is an interaction between a highly electro-negative element (F,O or N) and a hydrogen attached to a highly electro-negative element (F,O or N)

b) Dipole dipole interaction : the interaction is present in a molecule where bonds are polar [due to electro-negativity difference between bonded atoms]

c) Dispersion forces: these are weakest inter-molecular forces present between any two atoms due to development of temporary dipoles.

this is present in all molecules

In water  or H_{2}O: Hydrogen bonding, dipole–dipole, and dispersion

Where hydrogen is attached to oxygen and can form hydrogen bond

In CO  and CH3Cl: Dipole–dipole and dispersion only

In case of methane: there is no polar bond, no highly electronegative element so they have only Dispersion forces

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675 g of carbon tetrabromide is equivalent to how many
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<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
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<u>Chemistry</u>

<u>Organic</u>

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<u>Atomic Structure</u>

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<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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