Answer:
Mole fraction of solute is 0.0462
Explanation:
To solve this we use the colligative property of lowering vapor pressure.
First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr
Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.
Formula for lowering vapor pressure is:
ΔP = P° . Xm
Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)
Xm = mole fraction
24.9 mmHg - 23.8 mmHg = 23mmHg . Xm
Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg
Xm = 0.0462
In a reduction-oxidation or better known as REDOX reaction, the substance that reduces the oxidation state is known as the substance that is REDUCED. It serves as the oxidizing agent. Thus, Au3+ in this number is considered as the oxidizing agent.
Answer:
1.395J/g°C
Explanation:
The following were obtained from the question:
Q = 6527J
M = 312g
ΔT = 15°C
C =?
Q = MCΔT
C = Q/MΔT
C = 6527/(312 x 15)
C = 1.395J/g°C
The specific heat capacity of the substance is 1.395J/g°C
Answer:
3.43×10¹ mol
Explanation:
Given data:
Initial number of moles = 12.4 mol
Initial volume = 122.8 L
Final number of moles = ?
Final volume = 339.2 L
Solution:
The number of moles and volume are directly proportional to each other at same temperature and pressure.
V₁/n₁ = V₂/n₂
122.8 L/ 12.4 mol = 339.2 L / n₂
n₂ = 339.2 L× 12.4 mol / 122.8 L
n₂ = 4206.08 L.mol /122.8 L
n₂ = 34.3mol
In scientific notation:
3.43×10¹ mol
Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
