Answer:
is a combustion reaction
Explanation:
The reaction is a combustion since you have a hydrocarbon reacting with oxygen to form water and carbon dioxide.
The balanced equation is: 
If this helped, a brainliest answer would be greatly appreciated!
Answer:
0.1440M
Explanation:
Let''s bring out the parameters we were given;
Rate constant = 8.74 x 10^-4s^-1
Initial Concentration [A]o = 0.330M
Final concentration [A]= ?
Time = 800s
The reaction is a first order reaction, due to the unit of the rate constant. In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec.
Formular relating these parameters is given as;
ln[A] = ln[A]o − kt
Making [A] subject of interest, we have;
ln[A] = ln[A]o − kt
ln[A] = ln(0.330) - ( 8.74 x 10^-4 * 800)
In[A] = - 1.1086 - (6992 x 10^-4)
ln[A] = -1.8079
[A] = 0.1440M
You could look at it this way.
1 mole of solution = 0.132 of NaCl + x
1 - 0.132 = 0.868 mols of water.
answer: 0.868 <<<< answer
Answer:
the answer to your question is physical property
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
