The combustion of 30.0 g of glucose at room temperature and pressure produces 24.0 L of carbon dioxide.
<h3>What is combustion?</h3>
It is a reaction in which a substance burns with oxygen to form carbon dioxide and water.
Let's consider the combustion of glucose.
C₆H₁₂O₆ + 6 O₂ ⇒ 6 CO₂ + 6 H₂O
First, let's convert 30.0 g of glucose to moles using its molar mass.
30.0 g × 1 mol/180.16 g = 0.167 mol
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of carbon dioxide produced are:
0.167 mol Glucose × (6 mol CO₂/1 mol Glucose) = 1.00 mol CO₂
1 mol of an ideal gas at room temperature and pressure occupies 24.0 L.
The combustion of 30.0 g of glucose at room temperature and pressure produces 24.0 L of carbon dioxide.
Learn more about combustion here: brainly.com/question/9425444
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Answer:
C2H6 + 7/2O2 ....... 2CO2 + 3H2O
Answer:
u didn't give us the multiple hoice answer
Explanation:
Answer: Solubility.
Explanation:
Solubility is defined as the maximum amount of solute dissolved per 100 g of the solvent at a certain fixed temperature to form a saturated solution.
STP condition is Standard Temperature and Pressure condition which is temperature of 273 K and pressure of 1 atm.
Thus the scientific term for "the number of grams of solute dissolved in 100 g of the solvent to form a saturated solution at STP" is called as Solubility.
