Answer: The statement which could possibly not be true is C -" Liquid X can exist as a stable phase at 25°C, 1atm."
Explanation:
Triple point is the point where a substance co-exist as solid liquid and gas. At any point other than the triple point, the substance exist as a single phase substance.
As shown in the diagram, Liquid cannot exist as a stable phase at 1atm( below the the triple point pressure of 2atm) as the liquid can only exist beyond the pressure of triple point.
Answer:
The intermediates in this reaction are Cl and CCl₃.
Explanation:
- To indicate the intermediate in this reaction, we should firstly define the intermediate.
- The intermediate is the species that produced within the steps of the reaction and consumed in the later step/s and does not appear in the overall reaction (<em>neither reactants nor products</em>).
- The mechanism of the reaction contains 3 steps:
- Cl₂ ↔ 2Cl
- Cl + CHCl₃ → HCl + CCl₃
- Cl + CCl₃ → CCl₄
- The overall reaction is: Cl₂ + CHCl₃ → HCl + CCl₄
- So, the intermediates in this reaction are Cl and CCl₃.
- Thus, 2 moles of Cl is produced in the first step and consumed in the second and third steps.
- 1 mole of CCl₃ is produced in the second step and consumed in the third step.
Answer:
Potassium chloride
Explanation:
A solution is formed by a solvent and one or more solutes.
The solvent is the species that is in major proportion and usually defines the state of aggregation of the solution, while the solute/s is/are in minor proportion.
Also, water is known as the universal solvent, so in any solution containing water, it is considered as the solvent.
Then, in an aqueous solution of potassium chloride the solute is potassium chloride.
Answer:
26.0 moles
Explanation:
Given the formula;
PV =nRT
P= pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = gas constant
T = temperature
n = PV/RT
n = 125 * 5/0.082 * (20 + 273)
n = 625/24.026
n = 26.0 moles
Answer:
2.5 g
Explanation:
The partial coefficient (K) is the ratio of the solubility (S) of a solute in two different solvents. For the given problem:
K = Ssolvent/Swater
The solubility is the mass divided by the volume. If the fraction of the solute in water is called q, the fraction of it in the solvent is 1-q. Let's call m the initial mass of the solute, so, after the extraction:
mass at the solvent = (1-q)m
mass at water = qm
Ssolvent = (1-q)m/50
Swater = qm/80
8 = [(1-q)m/50]/qm/80
8 = 80*(1-q)/50q
400q = 80 - 80q
480q = 80
q = 0.167
The mass that remains at the water is:
0.167*15 = 2.5 g
