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
Answer:
for volume only liters can be used
Explanation:
Increases the volume of water
Explanation:
Freezing of water causes the volume of water to increase by a whooping 4%. This is why the density of water is slightly different that of ice.
When ice freezes it expands and takes up more volume of space for the same mass given.
This is why bottles break when water in them is frozen.
This increase in volume is why ice floats on water. It makes it less dense.
learn more:
Density brainly.com/question/3764212
#learnwithBrainly
Answer:
3 m/s
Explanation:
▲
usando la fórmula velocidad = frecuencia multiplicada por la longitud de onda.
v = (1.5)(2)
v= 3 m/s
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
