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:
C. 3.40 ppm, singlet
Explanation:
Given the information from the question .we have to select the best representation for represents the predicted approximate chemical shift and coupling for hydrogen(s). In this case, there is no neighboring hydrogen .Thus there won’t be any split .the best option answer is C. 3.40 ppm, singlet . Therefore the correct answer or option is C. 3.40 ppm, singlet.
Answer:
Period
Explanation:
horizontal is a period and verticle is a group
Is this a real question? It’s B
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.