Explanation:
It should be decreasing, and the potential energy increasing, since Law of Conversation of Energy, right?
Also, if you think about what happens when it goes down, it loses potential and gains kinetic, so maybe the opposite should happen when it goes up.
Answer:
T f = (Normal freezing point of benzene - Molality of the solution × freezing point depression per molal of benzene) = 5.5 ° C - .625 m × 5.12° C = 2.30° C.
T f = 2.30° C.
T b = (Normal boiling point of benzene - Molality of the solution × boiling point elevation per molal of benzene) = 80.1 ° C. - .625 × 2.53° C = 81.68 ° C.
T b = 81.68 ° C.
Explanation:
The molality of the solution is .541 ÷ .865 = .625 m.
Ideally benzene freezes at 5.5 ° C and the depression of the freezing point is 5.12° C per molal.
At a molality of 0.625 the freezing point will be depressed by (molality of solution × freezing point depression per molal of benzene) or .625 × 5.12° C = 3.20° C. So the resulting freezing point will be (Ideal freezing point - freezing point depression) or T f =5.5 ° C. - 3.20 ° C. = 2.30° C.
Also ideally benzene boils at a temperature of 80.1° C and the elevation of the boiling point per molal is 2.53° C . So the boiling point of benzene will be elevated by (molality of solution × boiling point elevation per molal of benzene) .625 × 2.53° C = 1.58° C. So the boiling point of the current solution, T b, will be 80.1 ° C. + 1.58 ° C. = 81.68 ° C.
Answer:
0.085 (Ms)⁻¹
Explanation:
Half life = 12 s
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
is the initial concentration = 0.98 M
Half life expression for second order kinetic is:
![t_{1/2}=\frac{1}{k[A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA_o%5D%7D)
So,
k = 0.085 (Ms)⁻¹
The rate constant for this reaction is <u>0.085 (Ms)⁻¹ .</u>
Standard enthalpy change of neutralization is the enthalpy change when 1 mole of water is produced by the reaction of an acid and a base under standard conditions.
Enthalpy change of neutralization:
- Every time enthalpy change of neutralization is calculated, one mole of water is produced.
- Heat is released when an acid and an alkali react, hence enthalpy changes of neutralization are always negative.
- The values are always quite comparable for reactions involving strong acids and alkalis, falling between -57 and -58 kJ mol-1.
- If the reaction is the same in each case of a strong acid and a strong alkali, the enthalpy change is similar.
Learn more about the Enthalpy of neutralization with the help of the given link:
brainly.com/question/15347368
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