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:
The correct answer is
[CH4][H2O]/[CO][H2]3
Option 3 is correct
Explanation:
Step 1: Data given
For the equation aA + bB ⇆ cC + dD
The equilibrium constant is [C]^c * [D]^d / [A]^a*[B]^b
Step 2: Calculate the equilibrium constant Kc
CO+3H2⇌CH4+H2O
Kc = [H2O][CH4] / [CO][H2]³
The correct answer is
[CH4][H2O]/[CO][H2]3
Option 3 is correct
Because the heat of the ocean mixed with the cold air above the ocean causes a hurricane
<span>Fermentation is a metabolic process that changes sugar to acids, gases, or alcohol. It happens in yeast and bacteria, and also in oxygen-starved muscle cells, as in the case of lactic acid fermentation.</span>
Mass of Oxygen (O₂) : = 88.16 g
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
C₆H₁₄+ O₂ → CO₂ + H₂0
25 g C₆H₁₄
Required
mass of oxygen (O₂)
Solution
Balanced equation
2C₆H₁₄ + 19O₂ ⇒12 CO₂ + 14 H₂O
mol C₆H₁₄ (MW=86,18 g/mol) :
= mass : MW
= 25 g : 86.18 g/mol
= 0.29
From the equation, mol ratio of C₆H₁₄ : O₂ = 2 : 19, so mol O₂ :
= 19/2 x mol C₆H₁₄
= 19/2 x 0.29
= 2.755
Mass O₂(MW=32 g/mol) :
= 2.755 x 32
= 88.16 g
