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675 g of carbon tetrabromide is equivalent to how many
mol. Make sure to round your answer to the nearest WHOLE NUMBER

Sagot :

Spaceidn

Answer:

2.04 mol CBr₄

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Organic

  • Writing Organic Compounds
  • Writing Covalent Compounds
  • Organic Prefixes

Atomic Structure

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

Step 1: Define

675 g CBr₄

Step 2: Identify Conversions

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

Step 3: Convert

[tex]\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

Sarah06109idn

Answer:

[tex]\boxed {\boxed {\sf About \ 2.04 \ moles \ of \ CBr_4}}[/tex]

Explanation:

1. Define Formula

The compound is carbon tetrabromide. The tetra indicates 4 atoms of bromine.

  • Carbon (C)+ 4 Bromine (Br) = CBr₄

2. Find Molar Mass

There are 2 elements in this compound: carbon and bromine. Use the Periodic Table to find the molar masses of these elements.

  • Carbon (C): 12.011 g/mol
  • Bromine (Br): 79.90 g/mol

The molar mass is based on the number of atoms in the compound. The compound has 1 atom of carbon and 4 atoms of bromine.

CBr₄= 1(12.011 g/mol) + 4(79.90 g/mol)

= 12.011 g/mol+319.60 g/mol = 331.611 g/mol

3. Convert Grams to Moles

We want to convert 675 grams to moles. We should use the molar mass as a fraction.

[tex]\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]

Multiply by the given number of grams.

[tex]675 \ g \ CBr_4 *\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]

Flip the fraction so the grams of CBr₄ will cancel.

[tex]675 \ g \ CBr_4 *\frac{1 \ mol \ CBr_4 }{331.611 \ g \ CBr_4}[/tex]= [tex]675 \ *\frac{1 \ mol \ CBr_4 }{331.611 \ }[/tex]

[tex]\frac{675 \ mol \ CBr_4 }{331.611 \ }[/tex] = [tex]2.03551752 \ mol \ CBr_4[/tex]

4.Round

The original measurement, 675 grams has 3 significant figures (6,7 and 5), so our answer must have the same.

For the answer we found, 3 sig figs is the hundredth place.

[tex]2.03551752 \ mol \ CBr_4[/tex]

The 5 in the thousandth place tells us to round the 3 to a 4.

[tex]\approx 2.04 \ mol \ CBr_4[/tex]

There are about 2.04 moles of carbon tetrabromide in 675 grams.

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