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
The answer is statement #3.
Explanation:
<h3 /><h2>
<em><u>H2 </u></em><em><u>+</u></em><em><u> </u></em><em><u>O2 </u></em><em><u>=</u></em><em><u> </u></em><em><u>H2O</u></em></h2>
<h2>
<em><u>Hydrogen</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>Oxygen</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Water</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>
<em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em>
Answer: I just took the test. The answer is D! (A single replacement reaction takes place because sodium is more reactive than hydrogen.)
Great question, but I believe you are mixing up atomic number with mass number. Assuming you are, 12.011 amu is the average mass of a carbon atom. For carbon, it can come in three forms: carbon-12, carbon-13, carbon-14. The number following carbon is the mass number of that particular carbon "isotope". The reason the average is so close to 12 is because carbon-12 is by far the most common, so the average should be (and is) very close to 12. Therefore, 12.011 is a weighted average of all carbon molecules, and carbon-14 is a particular carbon molecule that weighs 14 amu.
