Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products
Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.
For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.
Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.
Repeat the same process for C and D.
After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.
SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE
Answer:
3.24 × 10^5 J/mol
Explanation:
The activation energy of this reaction can be calculated using the equation:
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
Where; Ea = the activation energy (J/mol)
R = the ideal gas constant = 8.3145 J/Kmol
T1 and T2 = absolute temperatures (K)
k1 and k2 = the reaction rate constants at respective temperature
First, we need to convert the temperatures in °C to K
T(K) = T(°C) + 273.15
T1 = 325°C + 273.15
T1 = 598.15K
T2 = 407°C + 273.15
T2 = 680.15K
Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)
ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4
7.831 = Ea(2.417 × 10^-5)
Ea = 3.24 × 10^5 J/mol
Answer:
The mercury button battery line notation is:
Zn⁰| Zn²|| , HgO| Hg⁰||
Explanation:
Mercury batteries either use
pure mercury(II) oxide (HgO)—which is called mercuric oxide
or a mixture of HgO with manganese dioxide (MnO2) which is as the cathode.
Anode has two half reactions:
The first step is an electrochemical reaction step:
Zn + 4OH− → Zn(OH)4−2 + 2e−[3]
cathode has a half reaction :
HgO + H2O + 2e− → Hg + 2OH−[3]
+0.0977 V standard potential is of this reaction.
which is followed by a chemical reaction step:
Anode consists of oxidation:
Zn+2OH>ZnO+H2O+2e
which gives an overall anode half-reaction of:
Zn + 2OH− → ZnO + H2O + 2e−[3]
Therefore,
The overall reaction for the battery is:
Zn + HgO → ZnO + Hg
Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol
