Using the equilibrium constant expression and given concentrations of SO₂ and O₂, the approximate concentration of SO₃ at equilibrium is 0.436 M, rounded to 2 significant figures.
The equilibrium constant expression for the given reaction is:
Kc = [SO₃]² / ([SO₂]² * [O₂])
Given that the equilibrium constant (Kc) is 2.5 x 10³, [SO₂] = 0.0651 M, and [O₂] = 0.114 M, we can substitute these values into the equilibrium constant expression:
2.5 x 10³ = [SO₃]² / (0.0651² * 0.114)
To solve for [SO₃], we can rearrange the equation:
[SO₃]² = (2.5 x 10³) * (0.0651² * 0.114)
[SO₃]² = 0.1902643
Taking the square root of both sides:
[SO₃] = √(0.1902643)
[SO₃] ≈ 0.436 M
Therefore, at equilibrium, the concentration of SO₃ is approximately 0.436 M, rounded to 2 significant figures.
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Calculate the Kp for the following reaction at 25.0 °C:
H₂(g) + Br₂(g) 2 HBr (g)
Round your answer to 1 significant digit.
AG= -107
kJ
mol
The equilibrium constant for the reaction as it has been shown is [tex]5.7 * 10^{18}[/tex]
What is the equilibrium constant?The quantitative expression of the size of a chemical process at equilibrium is the equilibrium constant, abbreviated as K. It links the reactant and product concentrations (or partial pressures) in a chemical process and gives details on the make-up of the equilibrium mixture. It offers crucial details regarding the proportions of reactants and products.
We know that;
ΔG = -RTlnKp
Thus we have that;
Kp =[tex]e^-[/tex](ΔG/RT)
Kp = [tex]e^-[/tex](-107000 /8.314 * 298)
=[tex]5.7 * 10^{18}[/tex]
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if a given sample of metal has a mass of 2.68 g and a volume of 1.03 cm3, what is its density?
Answer: If a given sample of metal has a mass of 2.68 g and a volume of 1.03 cm³, the density of the metal will be 2.6019417476 g/cm³.
Explanation:
To find out the density of any object we must have known values of mass of the object and volume of the object.
Mass- Mass is the amount of matter present in any object or particle. The S.I. unit of mass is the kilogram.
Volume- Volume is defined as the amount of space occupied by an object or particle. The measuring unit of volume is cubic meter (m³)- for larger volumes and cubic centimeters (ccm³) and cubic millimeters (cmm³) for smaller volumes.
Density- Density is the measurement that compares the mass of an object with its volume. The S.I. unit of density is kilogram per cubic meter (kg /m³) and the C.G.S unit is gram per cubic centimeter ( g/ ccm³). Density is denoted by rho (ρ).
The density of an object can be calculated by the following formula:
Density (ρ) = mass (m)/ volume (v)
In the given question, the mass of the object is 2.68 g. i.e. m = 2.68 g and the volume of the given sample is 1.03 cm³ i.e. v = 1.03 cm³.
Hence, by using the above formula and putting the values of mass and volume, we can calculate the density of the sample as below-
Density = mass (m)/ volume (v)
= 2.68 / 1.03
= 2.6019417476 g/cm³
Therefore, for the given sample of metal that has a mass of 2.68 g and volume of 1.03 cm³ will have a density of 2.6019417476 g/cm³.
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2.000 grams of Tantalum (Ta) is allowed to combust inside a bomb calorimeter in an excess of O2. The temperature inside changes from 32.00 °C to 39.15 °C.
If the calorimeter constant is 1160 J/°C, what is the energy of formation of Ta2O5 in kJ/mol? (remember, it could be positive or negative).
You will first need to write the balanced chemical equation for the formation of Ta2O5 . Tantalum is stable in the solid state at 25 °C and 1.00 atm of pressure.
The energy of formation of [tex]Ta_2O_5[/tex] is -1198.47 kJ/mol.
2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]
1. Write the balanced chemical equation for the formation of [tex]Ta_2O_5[/tex]:
2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]
2. Calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 39.15 °C - 32.00 °C
ΔT = 7.15 °C
3. Convert the mass of Tantalum (Ta) to moles:
The molar mass of Tantalum (Ta) is 180.95 g/mol.
Moles of Ta = mass of Ta / molar mass of Ta
Moles of Ta = 2.000 g / 180.95 g/mol
Moles of Ta = 0.0110 mol
4. Calculate the energy change (ΔE) using the formula:
ΔE = q - CΔT
Where q is the heat absorbed or released, C is the calorimeter constant, and ΔT is the change in temperature.
5. Substitute the values into the formula:
ΔE = q - CΔT
ΔE = q - (1160 J/°C)(7.15 °C)
ΔE = q - 8294 J
6. The heat absorbed or released (q) can be calculated using the equation:
q = n × ΔH
Where n is the number of moles and ΔH is the molar enthalpy of the reaction.
7. Rearrange the equation to solve for ΔH:
ΔH = q / n
8. Convert the energy change (ΔE) to kilojoules:
1 kJ = 1000 J
ΔE = ΔE / 1000
9. Substitute the values into the equation:
ΔH = ΔE / n
ΔH = (-8294 J) / 0.0110 mol
ΔH = -753,090 J/mol
10. Convert the enthalpy change (ΔH) to kilojoules per mole:
ΔH = ΔH / 1000
ΔH = -753.09 kJ/mol
11. Since the stoichiometry of the balanced equation is 2:1, divide the enthalpy change by 2:
ΔH = -753.09 kJ/mol / 2
ΔH = -376.55 kJ/mol
12. The energy of formation of [tex]Ta_2O_5[/tex] is the negative of the enthalpy change:
Energy of formation = -ΔH
Energy of formation = -(-376.55 kJ/mol)
Energy of formation = 376.55 kJ/mol
13. Finally, round the answer to the appropriate number of significant figures:
Energy of formation of [tex]Ta_2O_5[/tex] = -1198.47 kJ/mol
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What is percent abundance of 18 medium nails 5 cm long?
From the attached image, the percentage abundance of 18 medium nails 5 cm long is 19%
Understanding Percentage AbundanceThe percent abundance refers to the proportion or percentage of a certain type or category within a given sample or population.
In the case of 18 medium nails that are 5 cm long, we have the information presented in the table and we do not need to do any mathematical calculations.
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A chemical reaction is run in which 691 Joules of heat are generated and the internal energy changes by -536 Joules.
Calculate w for the system.
w =
Joules
The work done for the system, given that the internal energy changes by -536 Joules, is 1227 joules
How do i determine the work done for the system?From the question given above, the following data were obtained:
Heat generated (q) = 691 JoulesChange in internal energy (ΔU) = -536 JoulesWork done (W) = ?The work done for the system can be obtained as illustrated below:
ΔU = q - w
Inputting the given parameters, we have:
-536 = 691 - w
Collect like terms
-536 - 691 = -w
-1227 = -w
Multiply through by -1
w = 1227 joules
Thus, we can conclude that the work done for the system is 1227 joules
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At a certain temperature it is found that 1.83 moles of H2, 2.33 moles of 02 and 3.95 moles of H2O are in equilibrium in a 8.1 L container according to the reaction below. What is the equilibrium constant?
2 H2 (g) + 02 (g) = 2 H20 (g)
Keep extra significant figures during the calculation and round your answer to 1 decimal place.
0.6 is the equilibrium constant for the given reaction.
To calculate the equilibrium constant (K) for the given reaction, we need to use the molar concentrations of the reactants and products at equilibrium. The equilibrium constant expression is given by:
[tex]K= [H_{2}O]^{2} / ([H_{2}^{2} * [O_{2}])[/tex]
Given the moles of H2, O2, and H2O in the 8.1 L container, we can convert them to molar concentrations by dividing the number of moles by the volume:
[H2] = 1.83 moles / 8.1 L
[O2] = 2.33 moles / 8.1 L
[H2O] = 3.95 moles / 8.1 L
Substituting these values into the equilibrium constant expression, we have:
K = [tex](3.95 / 8.1)^{2}[/tex] / ([tex](1.83 / 8.1)^{2}[/tex] * (2.33 / 8.1))
Evaluating this expression and rounding to one decimal place, we find the equilibrium constant to be:
K ≈ 0.6
Therefore, the equilibrium constant for the given reaction is approximately 0.6.
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A runner wants to run 12.8 km . She knows that her running pace is 6.5 mi/h . Part A How many minutes must she run?
answer and explanation :)
Which of the following statements is true?
A.
Chemical reactions can either absorb thermal energy or release thermal energy.
B.
Chemical reactions can only release thermal energy.
C.
Chemical reactions can only absorb thermal energy.
D.
Chemical reactions can neither absorb thermal energy nor release thermal energy.