Answer:
a. 5.39 atm
Explanation:
Pressure = ?
Volume = 1 L
Temperature = 298 K
Mass = 9.70g
The formular relating these variables is the ideal gas equation;
PV = nRT
where R = gas constant = 0.082057 L.atm.K-1.mol-1
n can be obtained from the formular below;
n = mass / molar mass = 9.70 / 44
n = 0.2205 mol
P = nRT / V
P = 0.2205 * 0.082057 * 298 / 1
P = 5.392 atm
The correct option is option A.
B. -1,+1,0
C. -2,+3,-1
D. 0,0,0
Three resonance structures contribute to the structure of dinitrogen monoxide.
The resonance structure is invoked when a single structure can not sufficiently explain all the bonding properties of a compound. All the various contributing structures contribute to the final structure of the compound but not all to the same degree.
There are three resonance structures of dinitrogen monoxide. The most stable structure is always the structure that has the formal charges as -1, +1 and zero as shown.
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Answer:
A. 0, +1, -1
Explanation:
You can draw the lewis structure for NNO 3 ways: With two double bonds N=N=O, with a triple bond between the N and O and single bond between the two N's, or a triple bond between the two N's and a single bond between the N and O.
The goal is to have formal charges that are as small as possible, to have no identical formal charges on adjacent atoms, and to have the most negative formal charge on the most electronegative atom. The most stable structure is the one with the triple bond between the two N's because it gives the formal charges 0, 1, and -1 respectively. Unlike the other two structures, the negative formal charge is correctly placed on O, the most electronegative atom.
An element is to an atomas an organ is to a cell. Just as atoms are the fundamental building blocks of elements, cells are the basic units of living organisms.
Elements are composed of atoms, each characterized by a specific number of protons, neutrons, and electrons.
Similarly, organs are composed of cells, each with specialized structures and functions that collectively contribute to the overall function of the organ.
The analogy highlights the hierarchical organization of matter and life, emphasizing how complex structures are formed from simpler components.
Just as elements combine to create diverse substances, cells come together to form intricate organs essential for life processes.
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Answer:
An element is to a compound as an organ is to tissue
Explanation:
Answer:
The empirical formula is C3H5
Explanation:
Step 1: Data given
Mass of the compound = 7.80 grams
Mass of CO2 = 25.1 grams
Molar mass of CO2 = 44.01 g/mol
Mass of H2O = 8.55 grams
Molar mass of H2O = 18.02 g/mol
Molar mass C = 12.01 g/mol
Molar mass H = 1.01 g/mol
Molar mass O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 25.1 grams / 44.01 g/mol
Moles CO2 = 0.570 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.570 moles CO2 we have 0.570 moles C
Step 4: Calculate mass C
Mass C = 0.570 moles * 12.01 g/mol
Mass C = 6.846 grams
Step 5: Calculate moles H2O
Moles H2O = 8.55 grams / 18.02 g/mol
Moles H2O = 0.474 moles
Step 6: Calculate moles H
For 1 mol H2O we have 2 moles H
For 0.474 moles H2O we have 2*0.474 = 0.948 moles H
Step 7: Calculate mass H
Mass H = 0.948 moles * 1.01 g/mol
Mass H = 0.957 grams
Step 8: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.570 moles / 0.570 = 1
H: 0.948 moles / 0.570 = 1.66
This means for 1 mol C we have 1.66 moles H OR for 3 moles C we have 5 moles H
The empirical formula is C3H5
To find the empirical formula of the hydrocarbon, divide the moles of CO2 and H2O by their molar masses. Use the smallest mole ratio to determine the empirical formula.
To find the empirical formula of the hydrocarbon, we need to determine the mole ratios between carbon and hydrogen in the compound. First, calculate the moles of CO2 produced by dividing the mass of CO2 by its molar mass. Next, calculate the moles of H2O produced by dividing the mass of H2O by its molar mass. Finally, divide the moles of each element by the smallest number of moles to obtain the mole ratio between carbon and hydrogen. The empirical formula is CnHm, where n and m represent the mole ratios of carbon and hydrogen, respectively.
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Answer:
a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K
b) At 753.55 ºC or higher
c )ΔG = 1.8 x 10⁴ J
K = 8.2 x 10⁻²
Explanation:
a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂
ΔHf kJ/mol -12.5 103.8 0
ΔGºf kJ/K 119.7 202.5 0
Sº J/mol 255 238 130.6*
Note: This value was not given in our question, but is necessary and can be found in standard handbooks.
Using Hess law to calculate ΔHºrxn we have
ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3
ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)
ΔHºrxn = 116.3 kJ
Similarly,
ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3
ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ
ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K
b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using
ΔGrxn = ΔHrxn -TΔS
we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.
0 = 116 kJ -T (0.113 kJ/K)
T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC
c) Again we will use
ΔGrxn = ΔHrxn -TΔS
to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.
ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K
ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )
Now for solving for K, the equation to use is
ΔG = -RTlnK and solve for K
- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)
K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²
The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.
The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.
The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.
With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.
For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.
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Answer:
To determine the value of Kp for the given equilibrium, we need to use the partial pressures of the gases involved.
In the balanced equation: 2 HI (g) ⇌ H₂ (g) + I₂ (g), the stoichiometric coefficients are 2, 1, and 1 respectively.
At equilibrium, the expression for Kp is given by:
Kp = (P(H₂) * P(I₂)) / (P(HI)²)
Using the provided partial pressures:
P(HI) = 1.9 atm
P(H₂) = 7.9 atm
P(I₂) = 2.3 atm
Substituting these values into the expression for Kp:
Kp = (7.9 * 2.3) / (1.9²)
Kp ≈ 19.5 / 3.61
Calculating the result:
Kp ≈ 5.4
Therefore, the value of Kp for the given equilibrium is approximately 5.4.
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