Answer:
a) 0 J
b) -2.67x10² J
c) -2.09x10³ J
Explanation:
For an isothermic expansion (with constant temperature) the work (W) is :
W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).
a) In vacuum, the relative pressure is 0 atm, so the work:
W = -0x(5.7 - 1.3)
W = 0 J
b) For a constant pressure of 0.60 atm
W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm
1 L.atm = 101.3 J
W = -2.64x101.3 = -2.67x10² J
c) For a pressure of 4.7 atm
W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L
1 atm.L = 101.3 J
W = -20.68x101.3 = -2.09x10³ J
The correct answer is that a compound that donates protons.
On the basis of Bronsted-Lowry concept, a compound which accepts proton is considered as a base and the compound which donates protons is considered to be an acid. The strong acids and bases get ionized completely in aqueous solution, while the weak acids and weak bases get ionize partially.
The conjugate base is illustrated as the species that is produced after the loss of proton of acid, while the conjugate acid refers to the species that is produced after the gain of protons.
Thus, the Bronsted-Lowry definition of an acid is a compound that donates protons.
Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.
Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.
Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.
Answer:
option C is the correct answer
