Explanation:
The reactants are Aluminium and Copper chloride.
The products are Copper and Aluminium chloride.
The correct answer is species
Answer:
Both roots are imaginary roots.
Explanation:
Consider these things:
If we try to solve x²+1 = 0, notice that we aren't able to solve the equation in Real Number system because there are no negative outputs for quadratic function.
Remember that quadratic function has range greater or equal to the max-min value.
x-axis plane represents the solutions of that equation. If a graph intersects x-axis plane then it has a solution.
While a graph that doesn't have any intersects on x-plane, it means that the equation for that graph doesn't have real solutions but imaginary solutions.
As you may notice some of parabola graph has one intersect, two intersects or none. One intersect is one solution to the equation — Two intersects are two solutions of the equation and lastly, no intersects mean that no real solutions and remain only imaginary solution.
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
Answer:
0.571 mol
Explanation:
Given data:
Number of moles of NaHCO₃ = 0.571 mol
Number of moles of CO₂ produced = ?
Solution:
Chemical equation:
NaHCO₃ + C₃H₆O₃ → CO₂ + C₃H₅NaO₃ + H₂O
Now we will compare the moles of CO₂ with NaHCO₃ from balance chemical equation.
NaHCO₃ : CO₂
1 : 1
0.571 : 0.571
So number of moles of CO₂ produced are 0.571.
