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xz_007 [3.2K]
3 years ago
12

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

2O were produced. In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
Chemistry
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

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Answer:

The new volume after the temperature reduced to -100 °C is 0.894 L

Explanation:

Step 1: Data given

Volume of nitrogen gas = 1.55 L

Temperature = 27.0 °C = 300 K

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Step 2: Calculate the new volume

V1/T1 = V2/T2

⇒with V1 = the initial volume of the gas = 1.55 L

⇒with T1 = the initial temperature = 300 K

⇒with V2 = the new volume = TO BE DETERMINED

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2 years ago
How many molecules are in 85g of silver nitrate?
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<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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
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<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

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[PT] Molar Mass of O - 16.00 g/mol

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

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
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<u>Step 4: Check</u>

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

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

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