Explanation:
Since, it is shown that the reaction has been reversed. Therefore, value of 
will become 
.
Hence, new 
= 
= 20
Also, the number of moles of each reactant has been halved. So, 
for the reaction 
will also get halved.
Therefore, = 
= 
= 4.47
As the value of 
is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.
As a result, will also increase with increase in temperature.
Answer:
Shown below
Explanation:
a) for BrN3
80+3(14)=122amu
b) forC2H6
2(12) + 6(1) = 30amu
C) for NF2
14+2(19) = 52amu
D) Al2S3
2(27) + 3(32)= 150amu
E) for Fe(NO3)3
56 + 3 [14+3(16)] =242amu
F) Mg3N2
3(24) + 2(14)= 100amu
G) for (NH4)2CO3
2[14 +4(1)] +12 +3(16)=96amu
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
