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
To begin calculating, there is one thing you need to remember :
Then we have
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>
I'm sure it will help.
PH = -log[H3O+]
Solving for [H3O+] gives
[H3O+] = 10^-pH
= 10^-3
or 1x10^-3 M
Answer:
possibly because of the malleability of metals
Explanation:
In metallic bonding, electrons are delocalized and move freely among nuclei. When a force is exerted n the metal, the nuclei shift, but the bonds do not break, giving metals their characteristic malleability.
Answer:
We use the following formula as given below Use the formula below to find the lone pair on the oxygen atom of the SO3 molecule. L.P (O) = V.E (O) – N.A (S-O) Lone pair on the terminal oxygen atom in SO3 = L.P (O)
Explanation:
