Answer:The change in concentration of a reactant or product per unit time
Explanation:
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Answer:</h3>
8CO₂
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Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Answer:
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Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g