Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
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Explanation:
The given question is incomplete, the complete question is:
A reaction between liquid reactants takes place at 10.0 °c in a sealed, evacuated vessel with a measured volume of 35.0 L. Measurements show that the reaction produced 28. g of dinitrogen difluoride gas. Calculate the pressure of dinitrogen difluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits. pressure:atm
Answer:
The correct answer is 0.28 atm.
Explanation:
The temperature mentioned in the given reaction is 10 degree C, which after conversion becomes 283 Kelvin (273+10 = 283K).
The volume mentioned in the reaction is 35 Liters.
The reaction produced 28 grams of dinitrogen difluoride gas (N2F2). The n or the no of moles of the gas can be determined with the help of the formula:
moles of N2F2 = mass/molar mass
= 28/66 (molar mass of N2F2 is 66 g/mol)
= 0.424
The pressure of the gas can be determined by using the equation of the ideal gas law, that is, PV = nRT
P * 35 = 0.424 * 0.0821 * 283
P = 0.28 atm
S-34 is stable. It will not decay. @ answers.com this is also a good site to find the answer to your questions.
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
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