Answer:
527.68 mL
Explanation:
We will assume that nitrogen is behaving as ideal gas here.
For ideal gas the gas law is:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
Where
P1= initial pressure = 740 torr
V1= initial volume = 500mL
T1= initial temperature = 25⁰C = 298 K
P2= final pressure = 760 torr
V2= final volume = ?
T2= final temperature = 50⁰C = 323 K
Putting values in the gas law
Final volume = [tex]\frac{740X500X323}{298X760}= 527.68 mL[/tex]
ASAP!! What are the equations that represent how the energy of an electromagnetic wave is related to the frequency of the wave and to the wavelength of the wave? Define h and c in your equations and give their values.
Answer:
E = h f (relation of energy with frequency)E = h c / λ (relation of energy with wavelength)h = Planck's constant ≈ 6.026 × 10⁻³⁴ J.sc = speed of light = 299,792,458 m/s in vacuum ≈ 3.0 × 10⁸ m/sExplanation:
The equation that represents how the energy of an electromagnetic wave is related to the frequency of the wave and to the wavelength of the wave is known as the Planck - Einstein equation.
Such equation states that the energy of a photon of light (electromagnetic radiation) is proportional to its frequency:
E = h fWhere:
E is the energy of the one photon h is the proportionality constant, named Planck's constant, and has value 6.62607004 × 10⁻³⁴ J.s.f is the frequency of the electromagnetic wave.Since the frequency and the wavelength are inversely related as per the equation λ = c / f, where λ is the wavelength and c is the speed of light, you can derive the equation that relates the energy of an electromagnetic wave with the wavelength:
E = h f ⇒ E = h c / λWhere λ is the wavelength and c is the speed of light.
The value of c in vacuum is a constant and is equal to 299,792,458 m/s m/s, which is usually approximated to 3.0 × 10⁸ m/s
*view screenshot for equations*
A particular reactant decomposes with a half‑life of 109 s when its initial concentration is 0.280 M. The same reactant decomposes with a half‑life of 231 s when its initial concentration is 0.132 M.
1. Determine the reaction order.
(A)1
(B)2
(C)0
2. What is the value and units of the rate constant for this reaction? ????=
Answer:
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
Explanation:
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Rate = [tex]\frac{1}{k[A_{0}]^{(n-1)}}[/tex]
Half life =[tex]K\frac{1}{[A_{0}]^{(n-1)} }[/tex]
[tex]\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }[/tex]
where n = order of reaction
Putting values
[tex]\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}[/tex]
[tex]0.472=(0.472)^{(n-1)}[/tex]
Hence n = 2
[tex]halflife=\frac{1}{k[A_{0}]}[/tex]
Putting values
[tex]231=\frac{1}{K(0.132)}[/tex]
K = 0.0328
The reaction order is 2 (option B), which is determined by observing the change in half-life with concentration. The rate constant for this second order reaction, calculated using the provided formula, is approximately 0.032 M^-1s^-1.
Explanation:The reaction order refers to how the rate of a chemical reaction is affected by the concentration of its reactants. When we see that the half-life of the reaction changes as the concentration of the reactant changes, this signifies that the order of the reaction is not zero. In zero order reactions, the half-life is dependent on the initial concentration of reactants. In this case, we see that the half-life of the reaction increases as the concentration decreases, which is characteristic of a second order reaction. Therefore, the answer to the first part is (B) 2.
For a second order reaction, the rate constant can be calculated using the formula t1/2 = 1 / [A]0k rearranging gives k = 1 / [A]0t1/2. Substituting for t1/2 = 109 s, and [A]0 = 0.280M, we get k = 1 / (0.280M * 109 s) which is approximately 0.032 M-1s-1. The units for the rate constant k for a second order reaction are M-1s-1.
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Standard temperature is exactly
a. 100ºC. b. 273ºC. c. 0ºC. d. 0 K.
Answer:
c. 0 °C
Explanation:
Standard temperature is 273.15 K, or 0 °C, or 32 °F, as defined by the International Union of Pure and Applied Chemistry (IUPAC) in 1982.
Answer:
Standard temperature is exactly (C) 0ºC
Explanation:
STP means standard temperature pressure at which
T = 0℃ or 273K and Pressure = 1 atm
NTP means Normal temperature pressure which means room temperature and pressure at which
T = 20℃ or 293 K and Pressure = 1 atm
SATP means standard Ambient temperature pressure at which
T = 25℃ or 298K and P = 1atm
In tropical countries, the climate will be warm and hence room temperature usually considered as
T = 25℃ or 298K and P = 1 atm
Lab temperature is same as that of room temperature.
A sample of NI3 is contained in a piston and cylinder. The samples rapidly decomposes to form nitrogen gas and iodine gas, and releases 3.30 kJ of heat and does 950 J of work.
What is change E?
Answer:
[tex]\boxed{\text{-4.25 kJ}}[/tex]
Explanation:
ΔE = q + w
By convention, anything leaving the system is negative and anything entering the system is positive.
Data:
q = -3.30 kJ
w = -950 J = -0.950 kJ
Calculation
ΔE = -3.30 - 0.950 = -4.25 kJ
[tex]\text{The change in E is }\boxed{\textbf{-4.25 kJ}}[/tex]
Final answer:
The change in internal energy (ΔE) for the decomposition of NI3 is -4250 J, which indicates the system lost energy.
Explanation:
The change in internal energy (ΔE) of the system can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system (q) minus the work done by the system (w): ΔE = q - w. In this scenario, the sample of NI3 decomposes, releasing heat (q = -3.30 kJ, since the system loses heat) and doing work (w = 950 J).
First, it is important to make sure both q and w are in the same units. We can convert kilojoules to joules by multiplying by 1000: -3.30 kJ = -3300 J.
Therefore, the change in internal energy ΔE is:
ΔE = q - w
ΔE = (-3300 J) - (950 J)
ΔE = -4250 J
The negative sign indicates that the total energy of the system decreased by 4250 joules.
Which factor is needed to begin a nuclear fission reaction?
high temperature
stable nuclei
addition of a neutron
low pressure
Answer:
addition of a neutron
Explanation:
Nuclear fission is a radioactive decay process in which a heavy nucleus spontaneously disintegrates into lighter ones accompanied by the release of energy.
An atom whose neutron/proton ratio is the same as that of the stability ratio of that atom is said to be stable. When such an atom is bombarded with a neutron particle, the stability ratio is offset. What results is a radioactive decay of such a nucleus. This form of decay is a nuclear fission.
A series of chain reaction is produced though this until the stability ratio is reached.
A compound has a percent composition of 54.5% carbon, 9.3% hydrogen and 36.2 % oxygen.If its molar mass is 88 g/mol, what is its molecular formula?
Complete the following:
Answer:
C₄H₈O₂.
Explanation:
Firstly, we can calculate the no. of moles (n) of each component using the relation:n = mass/atomic mass,
mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.
mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.
mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.
To get the empirical formula, we divide by the lowest no. of moles (2.26 mol) of O:
∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.
∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) = 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol) = 44.0 g/mol.
∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.
∴ The molecular formula is, 2(C₂H₄O), that is; (C₄H₈O₂)
At 40.8C the value of Kw is 2.92 3 10214. a. Calculate the [H1] and [OH2] in pure water at 40.8C. b. What is the pH of pure water at 40.8C? c. If the hydroxide ion concentration in a solution is 0.10 M, what is the pH at 40.8C?
Answer :
(a) The concentration of hydrogen ion and hydroxide ion are, [tex]1.708\times 10^{-7}M[/tex].
(b) The pH of pure water is, 6.78
(c) The pH of solution is, 13
Explanation :
(a) First we have to calculate the concentration of hydrogen ion and hydroxide ion.
As we know that,
[tex]K_w=[H^+][OH^-][/tex]
In pure water, the concentration of hydrogen ion and hydroxide ion are equal. So, let the concentration of hydroxide ion and hydrogen ion be, 'x'.
[tex]2.92\times 10^{-14}=(x)\times (x)[/tex]
[tex]2.92\times 10^{-14}=(x)^2[/tex]
[tex]x=1.708\times 10^{-7}M[/tex]
The concentration of hydrogen ion and hydroxide ion are, [tex]1.708\times 10^{-7}M[/tex].
(b) Now we have to calculate the pH of pure water.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (1.708\times 10^{-7})[/tex]
[tex]pH=6.78[/tex]
The pH of pure water is, 6.78
(c) In this, first we have to calculate the pOH when the concentration of hydroxide ion is, 0.10 M.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (0.10)[/tex]
[tex]pOH=1[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1=13[/tex]
The pH of solution is, 13
At 40.8°C, the [H+] and [OH-] concentrations in pure water are both 1.71 × 10^−7 M. The pH of pure water at 40.8°C is 6.77. If the [OH-] concentration in a solution is 0.10 M, the pH at 40.8°C is 13.00.
Explanation:At 40.8°C, the value of Kw is given as 2.92 × 10^−14.
a. To calculate the [H+] and [OH-] concentrations in pure water at 40.8°C, we can use the fact that [H+] × [OH-] = Kw. Since water is neutral, the concentrations of [H+] and [OH-] are equal.
Using the given value of Kw, we have:
[H+] × [H+] = 2.92 × 10^−14
Solving for [H+], we find that [H+] = [OH-] = 1.71 × 10^−7 M.
b. The pH of pure water at 40.8°C can be calculated using the formula pH = -log[H+]. Substituting the value of [H+], we find that the pH is equal to 6.77.
c. If the hydroxide ion concentration in a solution is 0.10 M, we can calculate the pH using the formula pH = 14 - pOH. Since the [OH-] concentration is given as 0.10 M, the pOH can be calculated as -log(0.10) = 1.00. Substituting this value into the pH formula, we find that the pH is equal to 13.00.
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With thermodynamics, one cannot determine ________. With thermodynamics, one cannot determine ________. the temperature at which a reaction will be spontaneous the value of the equilibrium constant the extent of a reaction the direction of a spontaneous reaction the speed of a reaction
Answer:
With thermodynamics, one cannot determine the speed of a reaction.
Explanation:
Chemical thermodynamics, a branch of chemistry that deals with study of interrelation of the heat and the work with the chemical reactions or with the physical changes of the state within confines of laws of thermodynamics.
Chemical thermodynamics' structure is derived from first two laws of chemical thermodynamics. From fundamental equations of Gibbs, a multitude of some equations which relates thermodynamic properties of thermodynamic system can be derived and can be used to calculate the reaction spontaneity, equilibrium constant, etc.
Thermodynamics predicts about the direction, feasibility and the extent of a chemical process, it does not tell anything about the rate at which a chemical process may proceed.
An object has a mass of 26.94 grams and a volume of 2.568 cubic centimeters. What material is it likely to be made
of?
Substance
iron
gold
silver
copper
Density (g/cm²)
7.874
19.32
110.49
20.93
Iron
Gold
Silver
Copper
Answer:
Explanation:
Formula
density = mass / volume
Givens
mass = 26.94
volume = 2.568
Solution
density = 26.94 / 2.568
density = 10.49 grams / cc^3
Answer
This is very likely silver.
The givens in the table except for the first 2 are wrong.
silver is as above
copper is a bit over 8.
Answer: The given material is silver.
Explanation:
To calculate density of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Mass of object = 26.94 g
Volume of object = [tex]2.568cm^3[/tex]
Putting values in above equation, we get:
[tex]\text{Density of object}=\frac{26.94g}{2.568cm^3}=10.49g/cm^3[/tex]
The calculated value of density corresponds to the density of silver.
Hence, the given material is silver.
A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.903 g of AgCl. Calculate the percent by mass of each compound in the mixture.
Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.
Explanation :
As we know that when a mixture of NaCl and KCl react with excess [tex]AgNO_3[/tex] then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.
Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.
The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.
First we have to calculate the moles of NaCl and KCl.
[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles[/tex]
[tex]\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles[/tex]
As, each mole of NaCl and KCl gives one mole of chloride ions.
So, moles of chloride ions in NaCl = [tex]\frac{x}{58.5}moles[/tex]
Moles of chloride ions in KCl = [tex]\frac{(0.9440-x)}{74.5}moles[/tex]
The total moles of chloride ions = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]
Now we have to calculate the moles of AgCl.
As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,
Moles of AgCl = Moles of chloride ion = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]
Now we have to calculate the moles of AgCl.
The molar mass of AgCl = 143.32 g/mole
[tex]\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles[/tex]
Now we have to determine the value of 'x'.
Moles of AgCl = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]
0.0133 mole = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]
By solving the term, we get the value of 'x'.
[tex]x=0.171g[/tex]
The mass of NaCl = x = 0.171 g
The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g
Now we have to calculate the mass percent of NaCl and KCl.
[tex]\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%[/tex]
[tex]\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%[/tex]
Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.
The combustion of titanium with oxygen produces titanium dioxide: Ti (s) + O2 (g) → TiO2 (s) When 2.060 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00°C to 91.60°C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is __________ kJ/mol.
Final answer:
The heat of reaction for the combustion of a mole of titanium in this calorimeter is 15220 kJ/mol.
Explanation:
In order to calculate the heat of reaction for the combustion of a mole of Ti in this calorimeter, we need to use the equation q = mcΔT. First, we need to calculate the amount of heat absorbed by the calorimeter and its contents.
The heat capacity of the calorimeter is given as 9.84 kJ/K, and the temperature change is 91.60°C - 25.00°C = 66.60°C. Therefore, the heat absorbed by the calorimeter and its contents is:
q = (9.84 kJ/K)(66.60°C) = 654.24 kJ.
Next, we need to determine the amount of moles in 2.060 g of titanium. The molar mass of titanium is 47.867 g/mol, so:
moles of titanium = 2.060 g / 47.867 g/mol = 0.043 moles.
Finally, we can calculate the heat of reaction for the combustion of a mole of Ti:
heat of reaction = 654.24 kJ / 0.043 moles = 15220 kJ/mol.
Use the reaction 2NO 2 (g) = N 2 O 4 (g) to answer the following question:
What does Le Châtelier's principle predict would happen to the equilibrium conditions if N2 O4 were added to the system?
Answer:
According to Le Chatelier’s Principle, a stress placed on a system at equilibrium will cause the equilibrium to shift to counteract the stress. For example, a temperature increase in the above reaction will favor the reverse reaction to use the excess heat and form brown NO2 gas. A temperature decrease in the above reaction favors the forward reaction to produce heat and form colorless N2O4 gas.
Answer:
The equilibrium shifts to the left and more of NO₂ is formed.
Explanation:
The Le Chatelier's principle states that when a form of stress is applied to a system in equilibrium, the system shifts so as to relieve that stress.
This principle explains that addition of one of the reactants to a system in equilibrium leads to the equilibrium shifting in such a way that the reaction occurring reduces the concentration of the added reactant.
In this case the addition of dinitrogen tetraoxide increases its concentration and therefore equilibrium shifts to reduce this concentration.
Which state of matter consists of particles that can be partially compressed? Gas Liquid Plasma Solid
Answer: Option (b) is the correct answer.
Explanation:
In liquid state, particles do have kinetic energy that helps in partially overcoming the intermolecular forces between the molecules. But still the particles are close together and they are able to slide past each other.
So, when we apply pressure on a liquid then its molecules partially gets compressed.
On the other hand, molecules of a solid are held together by strong intermolecular forces of attraction. Hence, they have definite shape and volume. As a result, solids do not get compressed.
In gases and plasma state of matter, molecules are gar away from each other. So, they are able to get completely compressed when a pressure is applied.
Thus, we can conclude that liquid is the state of matter which consists of particles that can be partially compressed.
Answer:
liquid
Explanation:
A sheet of gold weighing 11.4 g and at a temperature of 14.5°C is placed flat on a sheet of iron weighing 18.4 g and at a temperature of 55.4°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.
Answer:
The final temperature of the combined metals is 49.2314 °C
Explanation:
Heat gain by gold = Heat lost by iron
Thus,
[tex]m_{gold}\times C_{gold}\times (T_f-T_i)=-m_{iron}\times C_{iron}\times (T_f-T_i)[/tex]
Where, negative sign signifies heat loss
Or,
[tex]m_{gold}\times C_{gold}\times (T_f-T_i)=m_{iron}\times C_{iron}\times (T_i-T_f)[/tex]
For gold:
Mass = 11.4 g
Initial temperature = 14.5 °C
Specific heat of gold = 0.129 J/g°C
For iron:
Mass = 18.4 kg
Initial temperature = 55.4 °C
Specific heat of water = 0.450 J/g°C
So,
[tex]11.4\times 0.129\times (T_f-14.5)=18.4\times 0.450\times (55.4-T_f)[/tex]
[tex]1.4706\times (T_f-14.5)=8.28\times (55.4-T_f)[/tex]
[tex]1.4706\times T_f-1.4706\times 14.5=8.28\times 55.4-8.28\times T_f[/tex]
[tex]1.4706\times T_f-21.3237=458.712-8.28\times T_f[/tex]
[tex]1.4706\times T_f+8.28\times T_f=458.712+21.3237[/tex]
[tex]T_f=49.2314[/tex]
Thus,
The final temperature of the combined metals is 49.2314 °C
The final temperature of the combined metals is approximately 49.31°C.
Given data:
Mass of gold (m₁) = 11.4 gInitial temperature of gold (T₁) = 14.5°CMass of iron (m₂) = 18.4 gInitial temperature of iron (T₂) = 55.4°CSpecific heat capacity of gold (c₁) = 0.129 J/g°C (approximate value)Specific heat capacity of iron (c₂) = 0.449 J/g°C (approximate value)Step 1: Calculate the heat gained/lost by each metal
The heat gained by the gold (Q₁) is equal to the heat lost by the iron (Q₂), since no heat is lost to the surroundings.Q₁ = -Q₂The heat gained/lost by each metal can be calculated using the formula: Q = m × c × ΔT, where ΔT is the change in temperature.Step 2: Set up the equation
Let T_f be the final temperature of the combined metals. Then, the change in temperature for gold is T_f - T₁, and for iron, it's T_f - T₂.Q₁ = m₁ × c₁ × (T_f - T₁) Q₂ = m₂ × c₂ × (T_f - T₂)Since Q₁ = -Q₂, we can set up the equation: m₁ × c₁ × (T_f - T₁) = -m₂ × c₂ × (T_f - T₂)Step 3: Solve for T_f
Now, let's plug in the given values and solve for T_f: 11.4 g × 0.129 J/g°C × (T_f - 14.5°C) = -18.4 g × 0.449 J/g°C × (T_f - 55.4°C)Simplifying the equation, we get: 1.47 T_f - 21.23 = -8.23 T_f + 234.11Combine like terms: 9.70 T_f = 255.34Divide by 9.70: T_f = 255.34 / 9.70 T_f ≈ 49.31°CAnd there you have it! The final temperature of the combined metals is approximately 49.31°C.
What is the molar mass of (NH4)2C03?
a) 144 g
b) 96 g
c)138 g
d)78 g
Answer:
B
Explanation:
It has 2 NH4 molecules and 1 CO3 molecule.
NH4 has a molar mass of 18g/mol. Since there are two NH4s, it makes up 36g of the (NH4)2CO3.
CO3 has a molar mass of 60g/mol. Since there is one CO3, it makes up 60g of the (NH4)2CO3.
If you add up 36 and 60, you will get 96g.
A chemist has one solu6on that is 40% sulfuric acid and one that is 10% sulfuric acid. How much of each should she use to make 20 liters of a solu6on that is 28% sulfuric acid?
Answer:
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
Explanation:
For first solution of sulfuric acid :
C₁ = 40% , V₁ = ?
For second solution of sulfuric acid :
C₂ = 10% , V₂ = ?
For the resultant solution of sulfuric acid:
C₃ = 28% , V₃ = 20L
Also,
V₁ + V₂ = V₃ = 20L ......................................(1)
Using
C₁V₁ + C₂V₂ = C₃V₃
40×V₁ + 10×V₂ = 28×20
So,
40V₁ + 10V₂ = 560........................................(2)
Solving 1 and 2 as:
V₂ = 20 - V₁
Applying in 2
40V₁ + 10(20 - V₁) = 560
40V₁ + 200 - 10V₁ = 560
30V₁ = 360
V₁ = 12 L
So,
V₂ = 20 - V₁ = 8L
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess AgNO3, formed 2.044 g of AgCl. what is the percentage of NaCl in the impure sample?
Answer:
99.24%.
Explanation:
NaCl reacted with AgNO₃ as in the balanced equation:NaCl + AgNO₃ → AgCl(↓) + NaNO₃,
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
We need to calculate the no. of moles of AgCl produced:no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:using cross multiplication:
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
Now, we can get the mass of puree NaCl in the sample:mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.
Let's consider the reaction between NaCl and AgNO₃ to produce AgCl and NaNO₃.
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
We can calculate the mass of NaCl that produced 2.044 g of AgCl using the following relations.
The molar mass of AgCl is 143.32 g/mol.The molar ratio of AgCl to NaCl is 1:1.The molar mass of NaCl is 58.44 g/mol.[tex]2.044 g AgCl \times \frac{1molAgCl}{143.32 g AgCl} \times \frac{1molNaCl}{1molAgCl} \times \frac{58.44gNaCl}{1molNaCl} = 0.8335gNaCl[/tex]
An impure sample of mass 0.8421 g contains 0.8335 g of NaCl. The percentage of NaCl in the impure sample is:
[tex]\% NaCl = \frac{0.8335g}{0.8421g} \times 100\% = 98.98\%[/tex]
An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.
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When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from 23.40 °C to 24.21 °C. Calculate ΔHrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and C = 4.18 J/g ∙ °C as the specific heat capacity.
Answer : The enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, 67.716 KJ/mole
Explanation :
First we have to calculate the moles of [tex]AgNO_3[/tex] and [tex]HCl[/tex].
[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole[/tex]
[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole[/tex]
Now we have to calculate the moles of AgCl formed.
The balanced chemical reaction will be,
[tex]AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)[/tex]
As, 1 mole of [tex]AgNO_3[/tex] react with 1 mole of [tex]HCl[/tex] to give 1 mole of [tex]AgCl[/tex]
So, 0.005 mole of [tex]AgNO_3[/tex] react with 0.005 mole of [tex]HCl[/tex] to give 1 mole of [tex]AgCl[/tex]
The moles of AgCl formed = 0.005 mole
Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml
Now we have to calculate the mass of solution.
Mass of the solution = Density of the solution × Volume of the solution
Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g
Now we have to calculate the heat.
[tex]q=m\times C\Delta T=m\times C \times (T_2-T_1)[/tex]
where,
q = heat
C = specific heat capacity = [tex]4.18J/g^oC[/tex]
m = mass = 100 g
[tex]T_2[/tex] = final temperature = [tex]24.21^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]23.40^oC[/tex]
Now put all the given values in the above expression, we get:
[tex]q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC[/tex]
[tex]q=338.58J[/tex]
Now we have to calculate the enthalpy of the reaction.
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
[tex]\Delta H_{rxn}[/tex] = enthalpy of reaction = ?
q = heat of reaction = 338.58 J
n = moles of reaction = 0.005 mole
Now put all the given values in above expression, we get:
[tex]\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole[/tex]
Conversion used : (1 KJ = 1000 J)
Therefore, the enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, 67.716 KJ/mole
For this reaction, the heat of reaction is -67.8 KJ/mol.
The equation of the reaction is;
AgNO3(aq) + HCl(aq) -----> AgCl(s) + HNO3(aq)
Number of moles of AgNO3 = 50/1000 L × 0.100 M = 0.005 M
Number of moles of HCl = 50/1000 L × 0.100 M = 0.005 M
Temperature change = 24.21 °C - 23.40 °C = 0.81°C
Total volume of solution = 50.0 mL + 50.0 mL = 100 mL
Since the density of solution= 1.00 g/mL
Total mass of solution = 100g
Heat absorbed by solution = mcθ
m = mass of solution
c = specific heat capacity of solution
θ = temperature change
Heat absorbed by solution = 100g × 4.18 J/g ∙ °C × 0.81°C = 0.339 KJ
ΔHrxn = -( 0.339 KJ)/ 0.005 M
ΔHrxn = -67.8 KJ/mol
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An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a
An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a producer.
What are producer ?An organism that can make its own nourishment by the use of light, water, carbon dioxide, or other substances is an autotroph. Autotrophs are also referred to as producers because they make their own nourishment.
Producers are living things with the ability to grow their own nourishment. They frequently contain green vegetation. They use the process of photosynthesis to capture solar energy and use it to produce food. Since they are unable to create food on their own, all other creatures rely on producers for nourishment.
Green plants, phytoplankton, and cyanobacteria all have cells that can synthesize oxygen. Photosynthetic cells are highly diverse. Cells create sugar molecules and oxygen during the process of photosynthesis by using carbon dioxide and energy from the sun.
Thus, An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a producer.
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Enter a balanced chemical equation for the combustion of gaseous methanol.
The combustion reaction of methanol gaseous:
[tex]2CH_3OH(g) +3O_2(g) \longrightarrow 2CO_2(g) +4H_2O (g)[/tex]
What is the combustion reaction?A combustion reaction can be defined as a reaction that gives fire and takes place at an elevated temperature. A combustion reaction is an exothermic, redox reaction that commonly occurs between a hydrocarbon and mostly oxygen gas in the atmosphere.
Oxygen is the essential ingredient for the combustion reaction because combustion cannot happen without oxygen. An example of combustion is the burning of wood or solid fuels. The carbon in wood or coal reacts with oxygen gas in the air to liberate heat and gaseous products.
Combustion Reactions such as during the combustion of methanol react with molecules of oxygen gas to give carbon dioxide and water. The balanced chemical equation of the combustion reaction of methanol and oxygen is as follows:
[tex]2CH_3OH(g) +3O_2(g) \longrightarrow 2CO_2(g) +4H_2O (g)[/tex]
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H2(g) + Cl2(g) → 2HCl(g) ΔH = −184.6 kJ/mol. If 3.00 moles of H2 react with 3.00 moles of Cl2 to form HCl, what is ΔU (in kJ) for this reaction at 1.0 atm and 25°C? Assume the reaction goes to completion
Answer : The value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ
Explanation :
Formula used :
[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]
where,
[tex]\Delta E[/tex] = internal energy of the reaction = ?
[tex]\Delta H[/tex] = enthalpy of the reaction = -184.6 KJ/mole = -184600 J/mole
The balanced chemical reaction is,
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]
when the moles of [tex]H_2\text{ and }Cl_2[/tex] are 3 moles then the reaction will be,
[tex]3H_2(g)+3Cl_2(g)\rightarrow 6HCl(g)[/tex]
From the given balanced chemical reaction we conclude that,
[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 6 - 6 = 0 mole
R = gas constant = 8.314 J/mole.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta E=(-184600J/mole\times 3mole)-(0mole\times 8.314J/mole.K\times 298K)[/tex]
[tex]\Delta E=-553800J[/tex]
[tex]\Delta E=-553.8KJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ
The reaction H2(g) + Cl2(g) → 2HCl(g) is exothermic with a standard enthalpy change of -184.6 kJ/mol. For 3 moles each of H2 and Cl2, the energy released (ΔU) is -276.9 kJ. At 1 atm and 25°C, ΔH and ΔU are nearly the same due to insignificant work against atmospheric pressure.
Explanation:The given reaction (H2(g) + Cl2(g) → 2HCl(g)) is an exothermic reaction with a standard enthalpy change (ΔH) of -184.6 kJ/mol. This means that this amount of energy is released for every 1 mole of H2 and 1 mole of Cl2 that react to form 2 moles of HCl. If we're considering 3 moles of H2 and 3 moles Cl2, we are effectively dealing with 1.5 times the standard reaction. Hence, the total energy released (ΔU) is -184.6 kJ/mol * 1.5 = -276.9 kJ. An important note is that ΔH and ΔU are approximately the same for reactions at constant pressure and low temperatures (like 1 atm and 25°C), due to minimal work done against atmospheric pressure.
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Hydrocyanic acid is classified as a weak acid in water. This means that
A) it produces a relatively small fraction of the maximum number of possible hydronium ions.
B)100% of the maximum number of possible hydronium ions.
C)a relatively large fraction of the maximum number of possible hydronium ions.
D)no hydronium ions.
Answer:
A) it produces a relatively small fraction of the maximum number of possible hydronium ions.
Explanation:
A weak acid is one that ionizes slightly in aqueous solutions. They set up an equilibrium in the process. Weak acids like hydrocyanic acid only gives a very small fraction of the possible hydronium ions they ought to give in solution. They are not capable of ionizing completely like the strong acids. This makes hydrocyanic acid a weak electrolyte.
"The correct answer is A) it produces a relatively small fraction of the maximum number of possible hydronium ions. The correct statement about hydrocyanic acid is that it produces a relatively small fraction of the maximum number of possible hydronium ions when dissolved in water, which aligns with option A.
Hydrocyanic acid (HCN) is indeed classified as a weak acid in water. The classification of an acid as weak or strong is based on its ability to dissociate in water. A weak acid, such as hydrocyanic acid, only partially dissociates into its ions when dissolved in water. This means that only a small fraction of the HCN molecules ionize to release hydronium ions (H3O+) and cyanide ions (CN-).
In contrast, a strong acid would dissociate completely, producing the maximum number of hydronium ions possible from the acid. For example, hydrochloric acid (HCl) is a strong acid that dissociates completely in water to form H3O+ and Cl- ions.
The degree of dissociation of a weak acid is described by its acid dissociation constant (Ka), which is a measure of the strength of the acid in solution. The Ka value for hydrocyanic acid is relatively small, indicating that the equilibrium lies more towards the reactant side (undissociated HCN) rather than the product side (H3O+ and CN-).
Therefore, the correct statement about hydrocyanic acid is that it produces a relatively small fraction of the maximum number of possible hydronium ions when dissolved in water, which aligns with option A."
You are attempting to collect pure hydrogen gas by reacting hydrochloric acid with zinc. Based on the amount of reactants you used, you should have made 200 grams of hydrogen, but you only get 164 grams out of your reaction. What was your percent yield?
Answer: The percent yield of the given reaction is 82 %.
Explanation:
The chemical equation for the reaction of zinc and hydrochloric acid follows:
[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]
To calculate the percent yield of the reaction, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of hydrogen gas = 164 g
Theoretical yield of hydrogen gas = 200 g
Putting values in above equation, we get:
[tex]\%\text{ yield of hydrogen gas}=\frac{164g}{200g}\times 100\\\\\% \text{yield of hydrogen gas}=82\%[/tex]
Hence, the percent yield of the reaction is 82 %.
If it requires 38.5 milliliters of 0.85 molar nitric acid to neutralize 20.0 milliliters of barium hydroxide, solve for the molarity of barium hydroxide. Show all of the work used to solve this problem. (4 points) Unbalanced equation: Ba(OH)2 + HNO3 yields Ba(NO3)2 + H2O
Answer:
M of Ba(OH)₂ = 0.818 M.
Explanation:
The neutralization of H₂SO₄ with NaOH is occurred according to the balanced equation:Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O,
It is clear that every 1.0 mol of Ba(OH)₂ needs 2 mol of HNO₃ to be neutralized completely.
It is known at equivalence point: the no. of millimoles of base is equal to the no. of millimoles of acid.∴ (nMV) of Ba(OH)₂ = (nMV) for HNO₃.
where, n is the no. of producible H⁺ or OH⁻ of the acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
For Ba(OH)₂:
n = 2, M = ??? M, V = 20.0 mL.
For HNO₃:
n = 1, M = 0.85 M, V = 38.5 mL.
∴ M of Ba(OH)₂ = (nMV) of HNO₃ / (MV) for Ba(OH)₂ = (1)(0.85 M)(38.5 L)/(2)(20.0 mL) = 0.818 M.
You have two 466.0 mL aqueous solutions. Solution A is a solution of silver nitrate, and solution B is a solution of potassium chromate. The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g. (a) Calculate the concentration of the potassium ions in the original potassium chromate solution.(b) Calculate the concentration of the chromate ions in the final solution
Answer:
The concentration of the potassium ions in the original potassium chromate solution is 4.2927 mol/L.
The concentration of the chromate ions in the final solution is 1.0731 mol/L.
Explanation:
[tex]K_2CrO_4+2AgNO_3\rightarrow Ag_2CrO_4+2KNO_3[/tex]
Volume of solution A i.e. solution of silver nitrate = 466.0 mL = 0.466 L
Volume of solution B i.e. solution of potassium chromate = 466.0 mL = 0.466 L
Moles of silver chromate =[tex]\frac{331.8}{331.73 g/mol}=1.0002 mol[/tex]
According to reaction , 1 mol of silver chromate is produce from 2 moles of silver nitrate.
Then, 1.0002 moles of silver chromate will be formed from:
[tex]\frac{1}{2}\times 1.0002 mol=0.5001 mol[/tex] of silver nitrate.
According to reaction , 1 mol of silver chromate is produce from 1 mole of potassium chromate.
Then, 1.0002 moles of silver chromate will be formed from:
[tex]\frac{1}{1}\times 1.0002 mol=1.0002 mol[/tex] of potassium chromate
[tex]Concentration =\frac{Moles}{Volume (L)}[/tex]
a) The concentration of the potassium ions in the original potassium chromate solution.
Volume of the original solution = 0.466 L
1 mol of potassium chromate dissociates into 2mol of potassium ions and 1 mol of chromate ions:
Moles of potassium ions = 2 × 1.0002 mol = 2.0004 mol
[tex][K^+]=\frac{2.0004 mol}{0.466 L}=4.2927 mol/L[/tex]
b) The concentration of the chromate ions in the final solution
Volume of the final solution = 0.466 L + 0.466 L
Moles of chromate ions = 1 × 1.0002 mol = 1.0002 mol
[tex][CrO_4^{2+}]=\frac{1.0002 mol}{0.466 L+0.466L}=1.0731 mol/L[/tex]
Give a term for each description.Organic chemicals that can vaporize into the air.Secondary chemicals whose formation is facilitated by sunlight.Amount of gases and small particles in the atmosphere that influence ecosystem or human well-being.Gases or particles that are present in high enough concentrations to harm humans, other organisms, and buildings and other structures.Very small solid and liquid particles suspended in the air.
Answer:
Volatile organic compoundssecondary pollutants or photochemical substancesAir QualityAir pollutionAerosolsExplanation:
Volatile organic compounds are the organic chemicals that get easily vaporized in air.
The secondary chemicals or the secondary pollutants are the photochemical substances that are formed in the presence of sunlight.
The amount of gases and small particles present in the atmosphere, responsible for influencing ecosystem and the wellness of human beings is known as the Air Quality
Air pollution refers to the high concentrations of gases or small particles that are present in the atmosphere, which can cause harm to the humans and other living organisms and structures established by humans.
Aerosols are the tiny particles present in liquid or solid state, that are suspended in air.
If an aqueous solution of HNO3 is electrolyzed for 13.00 min at a steady current of 1.45 A, what volume of H2 (g) at 25.0oC and 1.03 atm will be collected at the cathode? vol H2 (L)
Answer: 0.14 Liters
Explanation:
[tex]Q=I\times t[/tex]
where Q= quantity of electricity in coloumbs
I = current in amperes = 1.45 A
t= time in seconds = 13 min=[tex]13\times 60 =780s[/tex]
[tex]Q=1.45A\times 780s=1131C[/tex]
[tex]HNO_3\rightarrow H^++NO_3^-[/tex]
[tex]2H^++2e^-\rightarrow H_2[/tex]
[tex]96500\times 2=193000Coloumb[/tex] of electricity deposits 1 mole of [tex]H_2[/tex]
1131 C of electricity deposits =[tex]\frac{1}{193000}\times 1131=5.86\times 10^{-3}moles[/tex] of [tex]H_2[/tex]
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1.03 atm
V= Volume of the gas = ?
T= Temperature of the gas = 25°C = 298 K (0°C = 273 K)
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= [tex]5.86\times 10^{-3}moles[/tex]
[tex]V=\frac{nRT}{P}=\frac{5.86\times 10^{-3}\times 0.0821\times 298}{1.03}=0.14L[/tex]
Thus the volume of hydrogen gas at [tex]25^0C[/tex] and 1.03 atm will be 0.14 Liters.
To calculate the volume of hydrogen gas produced at the cathode, Faraday's laws of electrolysis and the Ideal Gas Law are applied. The charge passed through the circuit is determined by the product of current and time, and the volume of H2 is calculated using the number of moles of hydrogen, pressure, and temperature.
Explanation:
Calculation of Hydrogen Gas Volume
To calculate the volume of hydrogen gas (H2) produced at the cathode during electrolysis, we use Faraday's laws of electrolysis which state that the amount of substance altered at an electrode during electrolysis is proportional to the amount of electricity that passes through the circuit. Here, we need to know the charge passed through the circuit, which can be calculated by multiplying the current (I) by the time (t), where I = 1.45 A and t = 13.00 min (converted to seconds).
The reaction at the cathode for the electrolysis of aqueous HNO3 will produce hydrogen gas according to the reaction:
2H2O(l) + 2e- → H2(g) + 2OH-
This means that 2 moles of electrons are required to produce 1 mole of H2. By using the Faraday constant (96,485 C/mol e-), we can calculate the moles of hydrogen produced:
Moles of electrons (n) = It/F, where F is the Faraday constant.
Finally, we can find the volume of H2 gas using the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of hydrogen, R is the Ideal Gas Constant (0.0821 L·atm/mol·K) and T is the temperature in Kelvin.
With the given temperature (25.0°C) and pressure (1.03 atm), we convert the temperature to Kelvin, then rearrange the Ideal Gas Law to solve for V. Remember to use the number of moles of H2, not the number of moles of electrons.
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of concentrated sulfuric acid if the density of the solution is 1.83 g cm-3.
Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Solution : Given,
Density of solution = [tex]1.83g/cm^3=1.83g/ml[/tex]
Molar mass of sulfuric acid (solute) = 98.079 g/mole
98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.
Mass of sulfuric acid (solute) = 98.0 g
Mass of solution = 100 g
Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g
First we have to calculate the volume of solution.
[tex]\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml[/tex]
Now we have to calculate the molarity of solution.
[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L[/tex]
Now we have to calculate the molality of the solution.
[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg[/tex]
Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
The molality of concentrated sulfuric acid is [tex]\( 0.500 \, \text{mol/kg} \)[/tex], and the molarity is [tex]\( 0.0183 \, \text{mol/L} \)[/tex].
1. Mass of sulfuric acid [tex](\( \text{H}_2\text{SO}_4 \))[/tex] in 100 g of solution:
[tex]\[ \text{Mass of } \text{H}_2\text{SO}_4 = 98.0 \% \times 100 \, \text{g} = 98.0 \, \text{g} \][/tex]
2. Moles of sulfuric acid [tex](\( \text{H}_2\text{SO}_4 \))[/tex]:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{98.0 \, \text{g}}{98.086 \, \text{g/mol}} \approx 1.000 \, \text{mol} \][/tex]
3. Molality [tex](\( m \))[/tex]:
[tex]\[ \text{Molality} = \frac{1.000 \, \text{mol}}{2.0 \, \text{kg}} = 0.500 \, \text{mol/kg} \][/tex]
4. Molarity [tex](\( M \))[/tex]:
[tex]\[ \text{Volume of solution} = \frac{100 \, \text{g}}{1.83 \times 10^{-3} \, \text{kg/cm}^3} \approx 54.6 \, \text{L} \]\[ \text{Molarity} = \frac{1.000 \, \text{mol}}{54.6 \, \text{L}} \approx 0.0183 \, \text{mol/L} \][/tex]
So, the molality of concentrated sulfuric acid is [tex]\( 0.500 \, \text{mol/kg} \)[/tex], and the molarity is [tex]\( 0.0183 \, \text{mol/L} \)[/tex].
Water has many exceptional and useful properties. Which is the rarest property among compounds?
Consider the following equation: N2O4(g) ⇄ 2 NO2(g) Kc = 5.8 × 10-3 If the initial concentration of N2O4(g) = 0.040 M and the initial concentration of NO2(g) is 0 M, what is the equilibrium concentration of N2O4(g)?
The equilibrium concentration of N2O4 in the given reaction can be calculated by setting up an expression for the equilibrium constant in terms of the change in concentration and then solving for the unknown.
Explanation:This question involves examining a chemical equilibrium problem for the reversible reaction: N2O4(g) ⇄ 2 NO2(g), with the given equilibrium constant, Kc = 5.8 × 10-3. The initial concentration of N2O4 is given as 0.040 M, while the initial concentration of NO2 is given as 0 M. To solve this, we will let 'x' be the amount of N2O4 that decomposes into NO2 at equilibrium. Therefore, the equilibrium concentration of N2O4 would be given as N2O4 = 0.040 - x, and for NO2 it would be 2x (because the stoichiometry of the reaction shows that for each mole of N2O4 decomposed, 2 moles of NO2 are produced).
Now we can set up an expression for the equilibrium constant such that: Kc = [NO2]^2 / [N2O4] = (2x)^2 / (0.040 - x). By substituting 5.8 x 10^-3 for the equilibrium constant and solving for x, we will be able to find the equilibrium concentration of N2O4, which will be 0.040 - x.
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