Answer:
–2733.4 KJ
Explanation:
The balanced equation for the reaction is given below:
C₂H₅OH + 3O₂ —> 2CO₂ + 3H₂O
ΔH = −1366.7 kJ
From the balanced equation above,
1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.
Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:
From the balanced equation above,
1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.
Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.
Thus, enthalpy change (ΔH) obtained is –2733.4 KJ
Answer:
E = 3.6×10⁻¹⁹ J
Explanation:
Given data:
Wavelength = 550 nm (550 ×10⁻⁹ nm)
Energy of wave = ?
Solution:
Formula:
E = h c/λ
c = 3×10⁸ m/s
h = 6.63×10⁻³⁴ Js
Now we will put the values in formula.
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s /550 ×10⁻⁹ nm
E = 19.89×10⁻²⁶ J.m /550 ×10⁻⁹ nm
E = 0.036×10⁻¹⁷ J
E = 3.6×10⁻¹⁹ J
Calcium chloride contains only one calcium and 2 chloride atom thus it's ionic formula would be CaCl2.
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺
+ 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu
⇄ Cu²⁺
+ 2e⁻
Ag⁺
+ e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺
+ 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu
+ 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu
+ 2Ag⁺
⇄ Cu²⁺
+ 2Ag
D is the answer. A, B,C are absurd. Doing a little word right there.