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Natasha_Volkova [10]
3 years ago
8

Two beakers are placed in a closed container. Beaker A initially contains 0.15 mole of naphthalene (C10H8) in 100 g of benzene (

C6H6), and beaker B initially contains 31 g of an unknown compound dissolved in 100 g of benzene. At equilibrium, beaker A is found to have lost 7.0 g of benzene. Assuming ideal behavior, calculate the molar mass of the unknown compound. State any assumptions made.
Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

Molar mass of the unknown compound is 192.2 g/mol

Explanation:

Assumption:

At equilibrium, molality of beaker A = molality of beaker B

Molality of beaker A = moles of naphthalene/kg of benzene

Mass of benzene = 100 - 7 = 93 g = 93/1000 = 0.093 kg

Molality of beaker A = 0.15/0.093 = 1.613 m

Molality of beaker B = moles of unknown compound (y)/kg of benzene

Mass of benzene = 100 g = 100/1000 = 0.1 kg

Molality of beaker B = (y/0.1) m

y/0.1 = 1.613

y = 0.1×1.613 = 0.1613 mol of unknown compound

Molar mass of unknown compound = mass/number of moles = 31 g/ 0.1613 mol = 192.2 g/mol

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Setler79 [48]

Answer:

\boxed {\boxed {\sf 53, 346 \ Joules}}

Explanation:

We are given the specific heat and change in temperature, so we should use this heat formula:

q=m C \Delta T

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

We know the mass is 150 grams. The specific heat of water is 4.184 J/g °C.

Let's find the change in temperature.

Subtract the initial temperature from the final temperature.

  • ΔT= final temp - initial temp
  • final= 95.0 °C and  initial= 10.0 °C
  • ΔT= 95.0 °C - 10.0 °C= 85.0 °C

Now we know all the values:

m= 150 \ g \\C= 4.184 J/ g \  \textdegree C \\\Delta T= 85.0 \textdegree C

Substitute them into the formula.

q=(150 \ g) (4.184 \ J/g \ \textdegree C)(85.0 \textdegree C )

Multiply all three numbers together. Note that the grams (g) and degrees Celsius (°C) will cancel out. Joules (J) will be the only remaining unit.

q=(627.6 \ J/ \textdegree C) ( 85.0 \textdegree C)

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4 0
3 years ago
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laiz [17]
<h3>Answer:</h3>

150 g Si

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Use Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 3.2 × 10²⁴ atoms Si

[Solve] grams Si

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

Avogadro's Number

[PT] Molar Mass of Si - 28.09 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.2 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})(\frac{28.09 \ g \ Si}{1 \ mol \ Si})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 149.266 \ g \ Si

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>

149.266 g Si ≈ 150 g Si

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