P1/V1=P2/V2
1/22.4=4/x
X=4 multiple by 22.4
V2=89.6L
Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
A catalyst is when a chemical reaction occurs faster than normal.
The system is unaffected during a catalyst because both forward and reverse reactions are affected, meaning that quilibrium will occur faster nothing will change.
Hope it helped,
BioTeacher101
Answer:
1.346 v
Explanation:
1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:
(oxidation) 
→
E°=0.337 v
(reduction) 
→
E°=1.679 v
(overall) 
+8H^{+}_{(aq)}→
E°=1.342 v
2) Nernst Equation
Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:
![E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}](https://tex.z-dn.net/?f=E%3DE%5E%7B0%7D%20-%5Cfrac%7BRT%7D%7BnF%7DLn%5Cfrac%7B%5Bred%5D%7D%7B%5Box%5D%7D)
Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.
The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give
![E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346](https://tex.z-dn.net/?f=E%3D1.342%20-%5Cfrac%7B298.15%2A8.314%7D%7B6%2A96500%7DLn%5Cfrac%7B%5B.66%5D%7D%7B%5B1.69%5D%7D%3D1.346)
E=1.346
