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
Stardust atoms are heavier elements, the percentage of star mass in our body is much more impressive. Most hydrogen in our body floats around in the form of water .
Explanation:
300c = 1.8f+32= please calculate it yourself ☺
300c= k+273= 573k
Answer:
0.508 L of solution.
Explanation:
Always a safe bet to convert to moles:
Where n is moles, m is mass, and MM is molar mass.
Now remember the equation for concentration (molarity):
Where C is the concentration, n is moles, and V is volume.
To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:
Therefore we can make 0.508 L of solution.
