Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

To calculate the yield follow these steps:
<u>1. Mole ratio</u>

<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles


ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles

ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
Answer:
The central atom has 3 electron domains.
Explanation:
According to the Valence Shell electron pair repulsion theory (VSEPR) put forward by Gillespie and Nyholm in 1957, the shape of a molecule is determined by repulsion between all the electron pairs (electron domains) present in the valence shell.
The electron pairs or electron domains are known to position themselves as far apart in space as possible in order to minimize repulsions.
Hence, when the central atom of a molecule contains three electron domains, they are positioned at an angle of 120° from each other to minimize repulsions. Hence the answer.
Carboxylic acid...........
Answer:
True or False?
Explanation:
A. False and or sometimes.
B. False and or sometimes.
What is our main objective here??????????????????
<span>The </span>standard enthalpy of formation<span> <span>is defined as the change in </span></span>enthalpy<span> <span>when one mole of a substance in the </span></span>standard<span> <span>state (1 atm of pressure and temperature of 298.15
K) is </span></span>formed<span> <span>from its
pure elements under the same conditions.</span></span>