Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol
Answer: an invisible line around which an object rotates, or spins.
Explanation: //Give thanks(and or Brainliest) if helpful (≧▽≦)//
Answer:
101,37°C
Explanation:
Boiling point elevation is one of the colligative properties of matter. The formula is:
ΔT = kb×m <em>(1)</em>
Where:
ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-
kb is ebulloscopic constant (0,52°C/m)
And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):
203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>
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Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m
Replacing these values in (1):
X - 100°C = 0,52°C/m×2,64m
X - 100°C = 1,37°C
<em>X = 101,37°C</em>
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I hope it helps!
Answer:
See explanation
Explanation:
The first step in this reaction is a unimolecular reaction. It involves the formation of the carbocation. This is so because tertiary alkyl halides only undergo substitution by SN1 mechanism due to sterric crowding.
The second step in the reaction is bi molecular. In this step, the carbocation now combines with the OH^- to yield the alcohol.
Net equation of the reaction is;
(CH3)3CBr + OH^- -------> (CH3)3COH + Br^-
The intermediate here is the carbocation, (CH3)3C^+
Lead (II) chloride+ potassium nitrate
PbCI2+KNO3(B)
