Answer:
friction
Explanation:
since it has a high tempature the friction increases like blowing air in a furnace
Glucose is the simplest sugar and carbohydrate that provides energy. The simplified model of glucose (C₆H₁₂O₆) shows carbon, hydrogen, and oxygen atoms linked together.
<h3>What is glucose?</h3>
Glucose is an example of a carbohydrate macromolecule that is further classified as a monosaccharide. They are crystalline and fundamental units of carbohydrates.
The molecular formula of glucose is C₆H₁₂O₆ and the mass is 180.156 g/mol. It is an aldohexose that contains an aldehydic functional group. In its structure, there are six oxygen atoms, six carbon atoms, and twelve hydrogen atoms.
Therefore, the glucose molecule is composed of C, H, and O.
Learn more about glucose here:
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Molar mass H₂SO₄ = 98.079 g/mol
1 mol -------- 98.079 g
? mole ------ 0.0960 g
moles = 0.0960 * 1 / 98.079
= 0.0960 / 98.079
= 9.788 x 10⁻⁴ moles
hope this helps!
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
