Mass of CO₂ produced : 58.67 g
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
CS₂ + 3O₂ -------> CO₂ + 2SO₂
mol of CO₂ based on mol of O₂ as a limiting reactant(CS₂ as an excess reactant)
From the equation, mol ratio of mol CO₂ : mol O₂ = 1 : 3, so mol CO₂ :

mass CO₂ (MW= 44 g/mol) :

<span>You may already know that when you breathe in, your body takes in oxygen from the air. When you breathe out, your lungs expel carbon dioxide back into the air. But the breath you breathe out contains more than just carbon dioxide.</span>
When you exhale (breathe out), your breath also containsmoisture. Because your mouth and lungs are moist, each breath you exhale contains a little bit of water in the form of water vapor(the gas form of water).
For water to stay a gas in the form of water vapor, it needs enough energy to keep its molecules moving. Inside your lungs where it's nice and warm, this isn't a problem.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
The average atomic mass of her sample is 114.54 amu
Let the 1st isotope be A
Let the 2nd isotope be B
From the question given above, the following data were obtained:
- Abundance of isotope A (A%) = 59.34%
- Mass of isotope A = 113.6459 amu
- Mass of isotope B = 115.8488 amu
- Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
- Average atomic mass =?
The average atomic mass of the sample can be obtained as follow:

Thus, the average atomic mass of the sample is 114.54 amu
Learn more about isotope: brainly.com/question/25868336