An element 'X after reacting with acids liberates hydrogen gas and candisplace lead and mercury from their salt solutions. The m
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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
Norfenefrine (C₈H₁₁NO₂).
<h3>Further explanation</h3>We will solve a case related to one of the colligative properties, namely freezing point depression.
The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.
<u>Given:</u>
A mysterious white powder could be,
- powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
- cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
- codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
- norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
- fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.
When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.
<u>Question: </u>What is the identity of the white powder?
<u>The Process:</u>
Let us identify the solute, the solvent, initial, and final temperatures.
- The solute = the powder
- The solvent = ethanol
- The freezing point of the solvent = −114.6°C
- The freezing point of the solution = −115.5°C
Prepare masses of solutes and solvents.
- Mass of solute = 82 mg = 0.082 g
- Mass of solvent = density x volume, i.e.,
We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.
The molality formula is as follows:
Now we combine it with the formula of freezing point depression.
It is clear that we will determine the molar mass of the solute (denoted by Mr).
We enter all data into the formula.
We get
These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.
<h3>Learn more</h3>- The molality and mole fraction of water brainly.com/question/10861444
- About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
- About the solution as a homogeneous mixture brainly.com/question/637791
Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity
