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
3-pentanone to form 3-pentanol
Here 3-pentanone reacts with H2/Pt to given 3-pentanol
So here with one step we can convert given ketone to alcohol
The reaction will be
Answer: There are 
molecules present in 7.62 L of 
at 
and 722 torr.
Explanation:
Given : Volume = 7.62 L
Temperature = 
Pressure = 722 torr
1 torr = 0.00131579
Converting torr into atm as follows.
Therefore, using the ideal gas equation the number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
According to the mole concept, 1 mole of every substance contains 
atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.
Thus, we can conclude that there are molecules present in 7.62 L of at and 722 torr.
The correct answer is D. A solid has a high to very high melting point, great hardness , and poor electrical conduction, this is a covalent network solid. This is because of the large amount of energy required to break these bonds, the said properties are exhibited by these solids.
