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lora16 [44]
2 years ago
11

How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel

sius?
**I NEED ANSWERS STEP BY STEP PLEASE**
Chemistry
1 answer:
o-na [289]1 year ago
7 0

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in m^3.

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= \frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

Learn more about ideal gas here:

brainly.com/question/16552394

#SPJ1

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5 0
10 months ago
2Mg + O2 → 2MgO<br><br> If you are burning 5.8332 g of Mg, how many grams of MgO will this make?
LekaFEV [45]
<h3>Answer:</h3>

9.6724 g MgO

<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>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Mg + O₂ → 2MgO

[Given] 5.8332 g Mg

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

[RxN] 2 mol Mg = 2 mol MgO

Molar Mass of Mg - 24.31 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of MgO - 24.31 + 16.00 = 40.31 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 5.8332 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{2 \ mol \ MgO}{2 \ mol \ Mg})(\frac{40.31 \ g \ MgO}{1 \ mol \ MgO})
  2. Multiply/Divide:                                                                                               \displaystyle 9.67241 \ g \ MgO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

9.67241 g MgO ≈ 9.6724 g MgO

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Now, examine the structures of benzhydrol and fluorene. Both compounds contain the same number of carbons but have very differen
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Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. <em>However, </em>the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.

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