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 chemical formula for the compound containing 8.6 mol of sulfur and 3.42 mol of phosphorus is P₂S₅
<h3>How do I determine the formula of the compound?</h3>
From the question given above, the following data were obatined:
- Sulphur (S) = 8.6 moles
- Phosphorus (P) = 3.42 mole
- Chemical formula =?
The chemical formula of the compound can be obtained as follow:
Divide by their molar mass
S = 8.6 / 32 = 0.26875
P = 3.42 / 31 = 0.11032
Divide by the smallest
S = 0.26875 / 0.11032 = 2.44
P = 0.11032 / 0.11032 = 1
Multiply by 2 to express in whole number
S = 2.44 × 2 = 5
P = 1 × 2 = 2
Thus, the chemical formula is P₂S₅
Learn more about empirical formula:
brainly.com/question/9459553
#SPJ1
by making sure they are in the lowest ratio. by adding them to see if they total 100. by checking that they are whole-number multiples. by dividing them by the molar mass.
PH + pOH = 14.
So, subtract 3.73 from 14.
The pOH is 10.27 :)
I believe the answer is A. :)<span />
