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
<span>Here's your balanced chemical equation:
Pb(NO3)2</span><span> (aq) + K2SO4 (aq) </span>⇒<span> PbSO4 (s) + 2KNO3 (aq)
Therefore the precipitate formed is </span>lead(II) <span>sulfate.</span>
The correct answer is option 2. One
way of expressing concentration is by expressing it by parts per million or
ppm. It <span>is calculated by dividing the grams of solute and the grams of the solution, and the result is multiplied by 1 000 000. </span><span>Parts per million is equal to 1 milligram of solute per kilogram of
solution. </span>
Each natural element has a characteristic light spectrum that helps identify it in samples of unknown substances. Spectroscopy is the practice of examining spectra and comparing them to those of known elements. Using spectroscopy methods, scientists can identify pure substances or compounds and the elements in them.
