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vlabodo [156]
3 years ago
12

An eccentric chemistry professor (not mentioning any names) stops in every day and orders 200ml of Sumatran coffee at precisely

75.0°C. You then need to add enough milk at 5.00°C to drop the temperature of the coffee, initially at 95.0°C, to the ordered temperature.
Chemistry
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

68.97g of milk must be added

Explanation:

To solve this problem we need to use the equation:

Q = C*m*ΔT -Coffee cup calorimeter equation-

This equation relates mass and change in temperature with heat of solution, Q.

The energy that decreases in the milk is the same that increases in the milk. That is:

Q(milk) = Q(tea)

Using the equation:

4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)

Mass milk * 58.0°C = 200g * 20.0°C

Mass milk = 4000g°C / 58.0°C

Mass milk =

68.97g of milk must be added

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loris [4]
<h3>Answer:</h3>

427 g Mo

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

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

Avogadro's Number

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<u>Step 3: Convert</u>

<u />2.68 \cdot 10^{24} \ atoms \ Mo(\frac{1 \ mol \ Mo}{6.022 \cdot 10^{23} \ atoms \ Mo} )(\frac{95.94 \ g \ Mo}{1 \ mol \  Mo} ) = 426.966 g Mo

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

426.966 g Mo ≈ 427 g Mo

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