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sweet [91]
3 years ago
14

2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to

relieve his heartburn. Unfortunately, he felt reaction happened in his stomach and the product which is CO2 gas was produced. Assume the pressure in Matt’s stomach was 780 torr and the amount of the gas was 400mL. What was the volume of the gas after it was released outside of Matt’s body? Assume human body temperature is 40°C and room temperature is 25°C. The pressure of the room is 1 atm. 760 torr = 1atm
Chemistry
1 answer:
evablogger [386]3 years ago
3 0

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

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katen-ka-za [31]

Question: Baking a Cake Without Flour.

Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.

Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.  

Results: My control cake, which I cooked for 25 minutes, measured 4 cm high.  Eight out of ten tasters that I picked at random from the class found it to be an acceptable dessert. After 25 minutes of baking, my experimental cake was 1.5 cm high and all ten tasters refused to eat it because it was burnt to a crisp.

What did I learn?/Conclusion: Since the experimental cake burned, my results did not support my hypothesis.  I think that the cake burned because it had less mass, but cooked for the same amount of time.  I propose that the baking time be shortened in subsequent trials.

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7 0
2 years ago
What is the mass in grams of 1.00 x 10 24 atoms of Mn?
Alex17521 [72]
<h3>Answer:</h3>

91.2 g Mn

<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>

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

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

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

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                           \displaystyle 91.2321 \ g \ Mn

<u>Step 4: Check</u>

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

91.2321 g Mn ≈ 91.2 g Mn

7 0
3 years ago
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