Answer:
2C4H10 + 13O2 ----> 4CO2 + 10H2O
The coefficient of oxygen in the balanced equation is 13
Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
Answer:
<em>Hi Todoroki here!!! </em>
Explanation:
Chlorine has the electron configuration [Ne]3s 2 3p 5, with the seven electrons in the third and outermost shell acting as its valence electrons. Like all halogens, it is thus one electron short of a full octet, and is hence a strong oxidising agent, reacting with many elements in order to complete its outer shell.
<em>Your welcome!!</em>
Answer: A: Explaining natural events using facts and data AND B:Developing theories using many lines of evidence
Explanation: trust me.
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
