Explanation:
Monitor the temperature of the water with the thermometer. Stop heating the water once it nears the boiling point of 100 degrees Celsius. Add copper(II) sulfate and stir until the heated solution is saturated. When the solution is saturated, copper(II) sulfate will not dissolve anymore
Answer:

Explanation:
From the question we are told that:
Density of acetic acid 
Density of Water 
Generally the equation for Solution Density is mathematically given by




Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C
You can work out C
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC
∆T
Polyatomic gas: C
= 3R
∆U= nC
∆T
∆U= 28g x C
x (350K - 58.3K)
∆U = 28C
x 291.7
∆U = 10967.6 x C
Zero because there are so many electrons as protons.
Answer:
Boron and Aluminium
Explanation:
Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.