The toboggan, having initial kinetic energy, gains potential energy as it goes up the hill until it stops. The kinetic energy's conversion into potential energy is described by the equation 0.5 * m * v^2 = m * g * h. TThe toboggan will stop approximately 2.82 m high above the base.
Explanation:This is a question about kinematics, more specifically potential and kinetic energy conversion in the presence of gravity. In this scenario, the kinetic energy of the toboggan is being converted into potential energy as it climbs up the hill.
The main formula we will use is the energy conservation law: Kinetic Energy = Potential Energy, or 0.5*m*v^2 = m*g*h where:
m is the mass, v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), h is the height.Even though we do not know the mass of the toboggan, it cancels out in the equation. So the actual height could be evaluated as h = v^2 / (2*g), substituting the given v and g into this formula, h = (11.5)^2 / (2*9.8) ≈ 6.75 m.
The actual vertical height above the base is h' = h * sinθ, where θ is the angle of the hill. Substituting into the formula, h' = 6.75 * sin(25) ≈ 2.82 m. So, the toboggan will go approximately 2.82m high vertically above the base before stopping.
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The toboggan will ascend the hill to a height of approximately 6.75 meters before stopping. This solution was derived using the principles of conservation of energy and kinematics.
Explanation:This question can be solved using the principles of conservation of energy and kinematics in physics. The toboggan, at the bottom of the hill, possesses kinetic energy (KE) which can be calculated using the formula KE = 1/2 m×v², where 'm' is the mass of the toboggan and 'v' is its speed.
As the toboggan moves up the icy, frictionless hill, this kinetic energy is gradually converted to gravitational potential energy (PE = m×g×h, where 'm' is the mass, 'g' is the acceleration due to gravity, and 'h' is the height). At the point when the toboggan stops, all the kinetic energy has been converted into potential energy.
It's important to note that we don't actually need the mass of the toboggan for this calculation because it cancels out in the equation KE = PE. Rearranging the equation to solve for 'h' gives us h = v² / (2*g)
Substituting the given values (v = 11.5 m/s and g = 9.81 m/s²), we get a height of approximately 6.75 meters. Therefore the toboggan will go up the hill to a height of approximately 6.75 meters above the base before stopping.
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A solution contains some or all of the following ions: Sn4+, Ag+, and Pb2+. The solution is treated as described below: Test 1) Addition of 6 M HCl causes a precipitate to form. Test 2) Addition of H2S and 0.2 M HCl to the liquid remaining from Test 1 produces no reaction. What conclusions can be drawn from the results of these two tests?
Answer:
In the first test precipitates AgCl and PbCl2. In the second one there is SnCl4 and SnS2 that are very soluble, and there ir more SnCl4 that SnS2.
Explanation:
This problem is about the cualitity studies about ions. The acidity is a factor for this studies. The chlorides and sulfides groups are mostly solubles except Pb2+, Ag+ and Hg2+ for chlorides and Sr+2, Ba+2, Pb+2 y Hg+2 for sulfides.
In the first case we have a high concentration of HCl. It means that all ions reaction with HCl. In the second one there is no reaction because in the solution we have SnCl4 that is very soluble and SnS2 is very soluble too. There is more SnCl4 because for Le Chatelier if we add more reactive the balance tends to reactive.
Which of the following acids is the STRONGEST? The acid is followed by its Ka value. Which of the following acids is the STRONGEST? The acid is followed by its Ka value. HF, 3.5 × 10-4 HCOOH, 1.8 × 10-4 HClO2, 1.1 × 10-2 HCN, 4.9 × 10-10 HNO2, 4.6 × 10-4
Answer:
chlorous acid HClO₂
Explanation:
The Ka is the acidity constant and it tells you about the acidity strength of a compound. If the value of Ka is high the compound is a strong acid. If the value of Ka is low the compound is a weak acid.
The problem gives us the following compounds with the Ka values:
HF, 3.5 × 10⁻⁴
HCOOH, 1.8 × 10⁻⁴
HClO₂, 1.1 × 10⁻²
HCN, 4.9 × 10⁻¹⁰
HNO₂, 4.6 × 10⁻4
The chlorous acid HClO₂ have the highest Ka, 1.1 × 10⁻², so this one is the strongest acid.
The strength of an acid is determined by its Ka value. Among the given options: HF, HCOOH, HClO2, HCN, and HNO2, the strongest acid is HClO2 as it has the highest Ka value of 1.1 × 10-2.
Explanation:The acidity of a substance is determined by its Ka value, which is the acid dissociation constant. A higher Ka value indicates a stronger acid because it shows the acid dissociates more completely in solution. The substances you've listed are all acids and their respective Ka values are given. Here, the strongest acid would have the highest Ka value.
Looking at the given options: HF (Ka = 3.5 × 10-4), HCOOH (Ka = 1.8 × 10-4), HClO2 (Ka = 1.1 × 10-2), HCN (Ka = 4.9 × 10-10), and HNO2 (Ka = 4.6 × 10-4), the strongest acid based on their Ka values would be HClO2 as it has the highest Ka value of 1.1 × 10-2.
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A chef is making deluxe sandwiches for a special guest. They call for 3 slices of jalapeno cheddar cheese and 5 slices of honey ham. If he has 180 slices of each in the kitchen storage, how many sandwiches can he make before one of his ingredients run out.?
Answer:
The chef can be able to make 36 sandwiches.
Explanation:
3 slices of jalapeno cheddar cheese + 5 slices of honey ham → 1 sandwich
According to information, 3 slices of jalapeno cheddar cheese will combine with 5 slices of honey ham to give 1 sandwich.
180 slices of jalapeno cheddar cheese will combine with:
[tex]\frac{5}{3}\times 180=300 [/tex] slices of honey ham
But we are having only 180 slices of honey ham. tghe number of sandwiches will depend upon number of slices of honey ham.
180 slices of honey ham will combine with:
[tex]\frac{3}{5}\times 180=108 [/tex] slices of jalapeno cheddar cheese
From 5 slices of honey ham we can make 1 sandwich,then from 180 slices of hinry ham we will be able make:
[tex]\frac{1}{5}\times 180=36 sandwiches[/tex]
The chef can be able to make 36 sandwiches.
Consider the following reactions: A: Uranium-238 emits an alpha particle B: Plutonium- 239 emits an alpha particle C: thorium-239 emits a beta particle
a. Rank the resulting nucleus by atomic number, from highest to lowest
b. Rank the resulting nucleus by the number of neutrons, from most to least
Answer:
Rank the resulting by neutrons, from most to least
The vapor pressure of water is 23.76 mm Hg at 25°C. How many grams of urea, CH4N2O, a nonvolatile, nonelectrolyte (MW = 60.10 g/mol), must be added to 238.2 grams of water to reduce the vapor pressure to 23.22 mm Hg ? water = H2O = 18.02 g/mol.
Answer:
18.700 g
Explanation:
As the urea is a nonvolatile and nonelectrolyte solute, it will reduce the vapor pressure of the solution according to:
[tex]P_{vs} =P_{w} *x_{w}[/tex]
Where [tex]P_{vs}[/tex] is the vapor pressure of the solution, [tex]P_{w}[/tex] is the vapor pressure of the pure water, and [tex]x_{w}[/tex] is the molar fraction of water. This equation applies just for that kind of solutes and at low pressures (23.76 mmHg is a low pressure).
From the equation above lets calculate the water molar fraction:
[tex]23.22mmHg=23.76mmHg*x_{w}\\ x_{w}=\frac{23.22mmHg}{23.76mmHg}=0.977[/tex]
So, the molar fraction of the urea should be: [tex]x_{urea}=1-x_{w}=0.023[/tex]
Then, calculate the average molecular weight:
[tex]M=x_{w}*MW_{w}+x_{urea}*MW_{urea}\\ M=0.977*18.02+0.023*60.10=18.989[/tex]
The molar fraction of urea is:
[tex]0.023=\frac{X urea mol}{S solution moles}=\frac{x urea grams}{238.2+x (solution grams)}*\frac{1 urea mol}{60.10 g}*\frac{18.989 solution grams}{1 solution mol}[/tex]
Solving for x,
[tex]x=18.700g[/tex]
Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800˚C (1073 K) is 3.6 × 1023 m-3 . The atomic weight and density (at 800˚C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3 .
Answer:
The energy for vacancy formation in silver is 1.1 ev/atom
Explanation:
The total number of sites is equal to:
[tex]N=\frac{N_{A} \rho }{A}[/tex]
Where
NA = Avogadro´s number = 6.023x10²³atom/mol
A = atomic weight of silver = 107.9 g/mol
ρ = density of silver = 9.5 g/cm³
Replacing:
[tex]N=\frac{6.023x10^{23}*9.5 }{107.9} =5.3x10^{22} atom/cm^{3} =5.3x10^{28} m^{-3}[/tex]
The energy for vacancy is equal:
[tex]Q=-RTln(\frac{N_{v} }{N} )[/tex]
Where
R = 8.314 J/mol K = 8.614x10⁻⁵ev/atom K
T = 800°C = 1073 K
Nv = number of vacancy = 3.6x10²³m⁻³
Replacing:
[tex]Q=-8.614x10^{-5} *1073*ln(\frac{3.6x10^{23} }{5.3x10^{28} } )=1.1ev/atom[/tex]
Final answer:
The energy for vacancy formation in silver is 2.87 x 10^-19 J.
Explanation:
To calculate the energy for vacancy formation in silver, we can use the equation: E = kTln(N/V), where E is the energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, N is the number of vacancies, and V is the volume.
Given that the equilibrium number of vacancies is 3.6 x 10^23 m^-3, we can use the density of silver to calculate the volume. The density of silver at 800˚C is 9.5 g/cm^3, which is equivalent to 9.5 x 10^6 kg/m^3.
Therefore, the energy for vacancy formation in silver is:
E = (1.38 x 10^-23 J/K) x (1073 K) x ln(3.6 x 10^23 / (9.5 x 10^6)) = 2.87 x 10^-19 J.
Suppose 4.00 mol of an ideal gas undergoes a reversible isothermal expansion from volume V1 to volume V2 = 9V1 at temperature T = 240 K. Find (a) the work done by the gas and (b) the entropy change of the gas. (c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?
Answer :
(a) The work done by the gas on the surroundings is, 17537.016 J
(b) The entropy change of the gas is, 73.0709 J/K
(c) The entropy change of the gas is equal to zero.
Explanation:
(a) The expression used for work done in reversible isothermal expansion will be,
[tex]w=nRT\ln (\frac{V_2}{V_1})[/tex]
where,
w = work done = ?
n = number of moles of gas = 4 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 240 K
[tex]V_1[/tex] = initial volume of gas = [tex]V_1[/tex]
[tex]V_2[/tex] = final volume of gas = [tex]9V_1[/tex]
Now put all the given values in the above formula, we get:
[tex]w=4mole\times 8.314J/moleK\times 240K\times \ln (\frac{9V_1}{V_1})[/tex]
[tex]w=17537.016J[/tex]
The work done by the gas on the surroundings is, 17537.016 J
(b) Now we have to calculate the entropy change of the gas.
As per first law of thermodynamic,
[tex]\Delta U=q-w[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
[tex]q=w[/tex]
Thus, w = q = 17537.016 J
Formula used for entropy change:
[tex]\Delta S=\frac{q}{T}[/tex]
[tex]\Delta S=\frac{17537.016J}{240K}=73.0709J/K[/tex]
The entropy change of the gas is, 73.0709 J/K
(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.
As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0
So, from this we conclude that the entropy change of the gas must also be equal to zero.
Ethylene diamine tetra-acetic acid (EDTA) is a water-soluble compound that readily combines with metals, such as calcium, magnesium, and iron. The molecular formula for EDTA is C10N2O8H16. One EDTA molecule complexes (associates with) one metal atom. A factory produces an aqueous waste that contains 20 mg/L calcium and collects the waste in 44-gallon drums. What mass (g) of EDTA would need to be added to each drum to completely complex all of the calcium in the barrel? (1 gal = 3.785 L)
Answer: The mass of EDTA that would be needed is 24.3 grams.
Explanation:
We are given:
Concentration of [tex]Ca^{2+}[/tex] ions = 20 mg/L
Converting this into grams/ Liter, we use the conversion factor:
1 g = 1000 mg
So, [tex]\Rightarrow \frac{20mg}{L}\times {1g}{1000mg}=0.02g/L[/tex]
Now, we need to calculate the mass of calcium present in 44 gallons of drum.
Conversion factor used: 1 gallon = 3.785 L
So, 44 gallons = (44 × 3.785)L = 166.54 L
Calculating the mass of calcium ions in given amount of volume, we get:
In 1L of volume, the mass of calcium ions present are 0.02 g.
Thus, in 166.54 L of volume, the mass of calcium ions present will be = [tex]\frac{0.02g}{1L}\times 166.54L=3.3308g[/tex]
The chemical equation for the reaction of calcium ion with EDTA to form Ca[EDTA] complex follows:
[tex]EDTA+Ca^{2+}\rightarrow Ca[EDTA][/tex]
Molar mass of EDTA = 292.24 g/mol
Molar mass of [tex]Ca^{2+}[/tex] ion = 40 g/mol
By Stoichiometry of the reaction:
40 grams of calcium ions reacts with 292.24 grams of EDTA.
So, 3.3308 grams of calcium ions will react with = [tex]\frac{292.24g}{40g}\times 3.3308g=24.33g[/tex] of EDTA.
Hence, the mass of EDTA that would be needed is 24.3 grams.
If the tip of the syringe, "The Titrator", was not filled with NaOH before the initial volume reading was recorded, would the concentration of acetic acid in vinegar of that trial be greater than or less than the actual concentration? Please explain your answer.
Answer:
The concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
If the syringe tip was not filled with NaOH before recording the initial volume, the concentration of acetic acid calculated would be less than the actual concentration.
If the tip of the syringe was not filled with NaOH before the initial volume reading was recorded, the concentration of acetic acid in vinegar calculated from that trial would be less than the actual concentration. This is because the actual volume of NaOH dispensed during titration would be over-reported.
When the titration is performed, the volume of NaOH required to reach the equivalence point appears larger than it truly is, leading to a miscalculation of the moles of NaOH used. Consequently, this will result in the calculated concentration of acetic acid being lower than its true value.
In titration, ensuring that the titrant (in this case, NaOH) is ready and properly measured is crucial for accurate results. The initial volume reading must be correct to avoid errors in determining the volume of NaOH added, which directly affects the accuracy of the acetic acid concentration estimation.
True or False Titanium's corrosion resistance is so strong that even titanium with oxygen impurities does no reduction in corrosion resistance.
Answer :
true
Explanation:
when titanium react with oxygen it form TiO₂ Titanium oxide which is passive in nature it forms layer of TiO₂ when large amount of oxygen is passed through titanium at very high temperature it does not react with oxygen impurities for this reason titanium mostly used in aerospace and chemical industries
A glycosidic bond can join two monosaccharide molecules to form a disaccharide.(T/F)
Answer:
True
Explanation:
A disaccharide is a sugar and the general molecular formula of a disaccharide is C₁₂H₂₂O₁₁.
A disaccharide is formed when two monosachharide units are joined by a covalent bond called the glycosidic bond.
The glycosidic bond in a disaccharide is formed by dehydration reaction between the two monosachharide units. The removal of the water molecule results in the formation of the glycosidic linkage.
For example: maltose a disaccharide, is formed when two molecules of glucose are joined by a (1→4) glycosidic bond. As, the glycosidic bond is formed between the carbon 1 of one glucose unit and carbon 4 of another glucose unit.
Therefore, in a disaccharide the two monosaccharide units are joined by a glycosidic bond or linkage.
Therefore, the given statement is TRUE.
vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compare this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?
Answer:
q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.
Explanation:
1. Heat absorbed
q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ
2. Change in volume
V(water) = 0.018 L
pV = nRT
1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K
V = 30.62 L
ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L
3. Work done
w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm
w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ
4. Why the difference?
Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.
The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
An achiral hydrocarbon A of molecular formula C7H12 reacts with two equivalents of H2 in the presence of Pd-C to form CH3CH2CH2CH2CH(CH3)2. One oxidative cleavage product formed by the treatment of A with O3 is CH3COOH. Reaction of A with H2 and Lindlar catalyst forms B, and reaction of A with Na, NH3 forms C. Identify compounds A, B, and C. Be sure to answer all parts.
Answer:
A) 5-methylhex-2-yne
B) (2Z)-5-methylhex-2-ene
C) (2E)-5-methylhex-2-ene
Explanation:
The given compound must be alkyne as it is undergoing reduction with two equivalents of hydrogen molecule.
Also as it is giving acetic acid on oxidative ozonolysis, it must have triple bond after two carbons in the chain.
The structure of hydrocarbon formed after reduction will give us the structure of alkyne by these information.
Reaction with hydrogen molecule in presence of Lindlar's catalyst gives cis alkene.
Reaction with hydrogen molecule in presence of Na, ammonia gives trans alkene.
The structure of compound is shown in the figure
The achiral hydrocarbon (A) with a molecular formula C7H12 is identified as Hept-1-yne. Upon reacting with a Lindlar catalyst and H2, it forms Hept-1-ene (B). When Hept-1-yne is treated with Sodium in ammonia (Na, NH3), it forms 1-heptyne (C).
Explanation:The hydrocarbon A that has a molecular formula of C7H12 should be Hept-1-yne in light of the fact that Hept-1-yne upon hydrogenation, utilizing Pd-C as a catalyst and two equivalents of H2, forms 4-methyl hexane, which is exact to what was stated in the question. When Hept-1-yne is treated with ozone (O3), it gives two oxidative cleavage products, one being CH3COOH (acetic acid). The reaction gives us a clue about the presence of a triple bond at the end of the heptane chain. Hence, the structure of compound A (Hept-1-yne) is identified as CH3-(CH2)4-C≡CH.
Compound B can be identified as Hept-1-ene. This is due to the fact that Hept-1-yne, upon reacting with a Lindlar catalyst and H2, forms Hept-1-ene. This conversion is a result of the partial reduction of the triple bond to a double bond.Lastly, compound C can be identified as 1-heptyne. This is because when we treat Hept-1-yne with Sodium in ammonia (Na, NH3, it selectively reduces the triple bond to a trans double bond, a process known as dissolving metal reduction.
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Phosphorous pentachloride is used in the industrial preparation of many organic phosphorous compounds. Equation I shows its preparation from PCl3 and Cl2: (I) PCl3 (l) + Cl2(g) PCl5(s) Use equation II and III to calculate ∆Hrxs of equation I: (II) P4 (s) + 6 Cl2 (g) 4 PCl3 (l) ∆H = 1280 KJ (III) P4 (s) + 10 Cl2 (g) 4 PCl5 (s) ∆H = 1774 KJ
Answer:
The enthalpy of the reaction is -123.5 kJ.
Explanation:
[tex]P4 (s) + 6 Cl_2 (g)\rightarrow 4 PCl_3 (l) ,\Delta H_1 =-1280 kJ[/tex]..(1)
[tex]P4 (s) + 10 Cl_2 (g)\rightarrow 4 PCl_5 (l) ,\Delta H_2 =-1774 kJ[/tex]..(2)
[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=x[/tex]...(3)
(2) - (1)
[tex]4PCl_3 (l) + 4Cl_2(g)\rightarrow 4PCl_5(s),\Delta H_{rxn}=y[/tex]
Dividing equation by 4 we get (3)
[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=\frac{y}{4}[/tex]...(3)
[tex]\Delta H_{rxn}=y=(-1774 kJ)-(-1280 kJ)=-494 kJ[/tex]
[tex]\Delta H_{rxn}=x=\frac{y}{4}={-494 kJ}{4}=-123.5 kJ[/tex]
The enthalpy of the reaction is -123.5 kJ.
Excited lithium ions emit radiation at a wavelength of 670.8 nm in the visible range spectrum. Calculate the frequency of a photon of this radiation
Answer : The frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]
Explanation : Given,
Wavelength of the radiation = 670.8 nm
First we have to convert wavelength form 'nm' to 'm'.
Conversion used : [tex](1nm=10^{-9}m)[/tex]
So, the wavelength of the radiation = 670.8 nm = [tex]670.8\times 10^{-9}m[/tex]
Now we have to calculate the frequency of a photon of radiation.
Formula used : [tex]\nu =\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of a photon of radiation
[tex]\lambda[/tex] = wavelength of the radiation
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get the frequency of a photon of radiation.
[tex]\nu =\frac{3\times 10^8m/s}{670.8\times 10^{-9}m}[/tex]
[tex]\nu =4.47\times 10^9s^{-1}[/tex]
Therefore, the frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]
Answer:
[tex]f=4.47x10^5GHz[/tex]
Explanation:
Hello,
In this case, we relate the speed of light, wavelength and frequency via the shown below equation expressed in the proper SI system of units:
[tex]f=\frac{c}{\lambda } =\frac{3x10^8m/s}{670.8nm*\frac{1x10^{-9}m}{1nm} } =4.47x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ f=4.47x10^5GHz[/tex]
Best regards.
Why is the combined cycle power generation system so much more efficient that the straight steam cycle?
Answer:
Because it uses the residual energy of the fluid used by the first engine.
Explanation:
A combined cycle power generation counts with two heat engines that work in tandem from the same source of heat. The engines turn the energy into mechanical energy.
The cycle is much more efficient than the other, almost 60% more.
I hope this answer helps you.
C2H4(g) + H2(g) → C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K Calculate ΔG at 25 °C and determine whether the reaction is spontaneous. Does ΔG become more negative or more positive as the temperature increases?
Answer:-ΔG=-101.5KJ
Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.
We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin
Also we need to convert the temperature in Kelvin as it is given in degree celsius.
[tex]\Delta H=-137.5\\ \Delta S=-120J/K\\ \Delta S=-0.120KJ/K\\ T=25^{.C}\\ T=273+25=298 K\\ \Delta G=?\\ \Delta G=\Delta H-T\Delta S\\\Delta G=-137.5KJ-(278\times -0.120)\\ \Delta G=-137.5+35.76\\\Delta G=-101.74\\\Delta G=Negative[/tex]
After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ
For a reaction to be spontaneous the value of \Delta G \ must be negative .
As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.
In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .
Hence the value of ΔG would become more positive with the increase in temperature.
So we found the value of ΔG to be -101.74KJ
Answer:
ΔG = -101.591 KJ
Explanation:
Gibbs free energy -
It is a thermodynamic quantity , which is given by the change in enthalpy minus the product of the change in entropy and absolute temperature.
i.e.,
ΔG is given as the change in gibbs free energy ( KJ )
ΔS is given as the change in entropy ( KJ /K )
ΔH is given as the change in ethalphy ( KJ )
T = temperature ( Kelvin ( K ))
ΔG = ΔH - TΔS
The sign of ΔG determines the reaction spontaneity , as
ΔG = negative , the reaction is spontaneous and
If ΔG = positive , the reaction is non spontaneous .
Given -
For the reaction ,
C₂H₄ (g) + H₂(g) ---> C₂H₆(g)
ΔH = - 137.5 KJ
ΔS = - 120.5 J /K
Since ,
1 KJ = 1000 J
1 J = 1 / 1000KJ
ΔS = - 120.5 / 1000 KJ /K
ΔS = -0.1205 KJ /K
T = 25°C
(adding 273 To °C to convert it to K)
T = 25 + 273 = 298 K
Putting the values on the above equation ,
ΔG = ΔH - TΔS
ΔG = -137.5 KJ - 298 * (-0.1205 KJ / K)
ΔG = -137.5 KJ + 35.909 KJ
ΔG = -101.591 KJ
Since,
the value of ΔG is negative ,
hence, the reaction is spontaneous.
For the above reaction ,
If the temperature is increased ,
ΔG = ΔH - TΔS
From the above equation ,
the value of TΔS will increase ,
As a result the value of ΔG will be more positive , by increasing the temperature.
A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from the nitrogen atoms and compound Y is produced from the metal atoms in the reactant. What are the compounds X and Y? X is nitrogen dioxide, and Y is a metal halide. X is nitrogen dioxide, and Y is a metal oxide. X is nitrogen gas, and Y is a metal sulfate. X is nitrogen gas, and Y is a metal oxide.
Answer:
The correct answer is: X is nitrogen dioxide, and Y is a metal oxide
Explanation:
Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.
The general equation is given as:
[tex]4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x[/tex]
Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.
In a combustion reaction, a compound containing nitrogen and a metal typically forms nitrogen dioxide (compound X) from the nitrogen atoms and a metal oxide (compound Y) from the metal atoms.
Explanation:In the context of your question about how a compound containing nitrogen and a metal reacts in a combustion reaction, the outcome depends on the specific reactant. Usually, compounds containing nitrogen atoms tend to form nitrogen oxides under high heat or combustion conditions, with nitrogen dioxide (NO2) being a common example. This would be compound X.
Regarding the metal component, in a combustion reaction, metals commonly react with oxygen in the environment to produce metal oxides. This would make compound Y a metal oxide. Therefore, the correct pair of products according to your question tends to be: X is nitrogen dioxide, and Y is a metal oxide.
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Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the following information: 1. 2CO2(g)+2H2O(l)⇌CH3COOH(l)+2O2(g), K1 = 5.40×10−16 2. 2H2(g)+O2(g)⇌2H2O(l), K2 = 1.06×1010 3. CH3COOH(l)⇌2C(s)+2H2(g)+O2(g), K3 = 2.68×10−9
Answer : The value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]
Explanation :
The following equilibrium reactions are :
(1) [tex]2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2[/tex] [tex]K_1=5.40\times 10^{-16}[/tex]
(2) [tex]2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)[/tex] [tex]K_2=1.06\times 10^{10}[/tex]
(3) [tex]CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)[/tex] [tex]K_3=2.68\times 10^{-9}[/tex]
The final equilibrium reaction is :
[tex]CO_2(g)\rightleftharpoons C(s)+O_2(g)[/tex] [tex]K_{goal}=?[/tex]
Now we have to calculate the value of [tex]K_{goal}[/tex] for the final reaction.
First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:
[tex]K_{goal}=\sqrt{K_1\times K_2\times K_3}[/tex]
Now put all the given values in this expression, we get :
[tex]K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}[/tex]
[tex]K_{goal}=1.238\times 10^{-7}[/tex]
Therefore, the value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]
The value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g) can be calculated by multiplying the equilibrium constants of the individual reactions involved.
Explanation:The value of the equilibrium constant, Kgoal, for the reaction CO2(g) ⇌ C(s) + O2(g) can be determined using the given information:
2CO2(g) + 2H2O(l) ⇌ CH3COOH(l) + 2O2(g), K1 = 5.40×10-162H2(g) + O2(g) ⇌ 2H2O(l), K2 = 1.06×1010CH3COOH(l) ⇌ 2C(s) + 2H2(g) + O2(g), K3 = 2.68×10-9Since reaction 3 is the sum of reactions 1 and 2, we can use the equations to calculate the value of Kgoal:
Kgoal = K1 × K2 × K3
Substituting the values:
Kgoal = (5.40×10-16) × (1.06×1010) × (2.68×10-9) = 1.47×10-14
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Calculate the equilibrium constants K’eq for each of the following reactions at pH 7.0 and 25oC, using the ∆Go’ values given: (a) Glucose-6-phosphate + H2O → glucose + PI ∆Go’= -13.8 kJ/mol (b) Lactose + H2O → glucose + galactose ∆Go’= -15.9 kJ/mol (c) Malate → fumarate + H2O ∆Go’= +3.1 kJ/mol
The equilibrium constants K'eq for the given reactions can be calculated using the formula K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin.
Explanation:For the given reactions the equilibrium constants K'eq can be calculated using the formula: K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol K at 25°C or 298.15 K) and T is the temperature in Kelvin.
For the reaction of Glucose-6-phosphate + H2O → glucose + PI with ∆Go'= -13.8 kJ/mol, K'eq = exp([(-13.8 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])For the reaction of Lactose + H2O → glucose + galactose with ∆Go'= -15.9 kJ/mol, K'eq = exp([(-15.9 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])For the reaction of Malate → fumarate + H2O with ∆Go'= +3.1 kJ/mol, K'eq = exp([(3.1 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])
Note: In the formulas above, the Gibbs free energy change is converted to J/mol (from kJ/mol) by multiplying by 10^3.
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The limiting reactant is completely consumed in a chemical reaction. (T/F)
Answer: Yes
Explanation:
Limiting reagent is the reagent which limits the formation of product as it gets completely consumed in the reaction.
Excess reagent is the reagent which is left unreacted in the reaction.
For example: [tex]2HCl+Ca\rightarrow CaCl_2+H_2[/tex]
If there are 2 moles of [tex]HCl[/tex] and 2 moles of [tex]Ca[/tex]
As can be seen from the chemical equation,
2 moles of hydrochloric acid react with 1 mole of calcium.
Thus 2 moles of [tex]HCl[/tex] will completely react with 1 mole of calcium and (2-1)=1 mole of calcium will remain as such.
Thus HCl is the limiting reagent as it limits the formation of product and calcium is the excess reagent as it is left unreacted.
Final answer:
The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction. It is completely consumed in the reaction.
Explanation:
The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. The other reactant or reactants are considered to be in excess. To determine the limiting reactant, you need to compare the amount of each reactant present in the reaction to the stoichiometric ratios in the balanced chemical equation. The reactant with the smallest amount is the limiting reactant.
In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL + 10.00 mL of the acid have been added?
Answer: The pH of the solution is 1.136
Explanation:
To calculate the moles from molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
For ammonia:Molarity of ammonia = 0.3764 M
Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol[/tex]
For nitric acid:Molarity of nitric acid = 0.3838 M
Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L
Putting values in above equation, we get:
[tex]0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol[/tex]
After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol
For the reaction of ammonia with nitric acid, the equation follows:
[tex]NH_3+HNO_3\rightarrow NH_4NO_3[/tex]
At [tex]t=0[/tex] 0.0178 0.022
Completion 0 0.0042 0.0178
As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.
To calculate the pH of the reaction, we use the equation:
[tex]pH=-\log[H^+][/tex]
where,
[tex][H^+]=\frac{0.0042mol}{0.05741L}=0.0731M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(0.0731)\\\\pH=1.136[/tex]
Hence, the pH of the solution is 1.136
Part C Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm. What is the radius of a gallium atom? Express your answer numerically in picometers. View Available Hint(s) radius = nothing pm p m Submit
Final answer:
The radius of a gallium atom in a primitive cubic unit cell, with an edge length of 362 pm, is calculated to be 181 picometers by dividing the edge length by 2.
Explanation:
The question asks for the radius of a gallium atom given that gallium crystallizes in a primitive cubic unit cell with an edge length of 362 pm. In a primitive cubic unit cell, the atoms are located at the corners of the cube, and the length of the edge of the cube is equal to twice the atomic radius. Therefore, to find the radius of the gallium atom, we divide the edge length by 2.
Radius of gallium atom = edge length / 2 = 362 pm / 2 = 181 pm.
This calculation reveals that the radius of a gallium atom is 181 picometers in a primitive cubic unit cell structure.
Enzymes function most efficiently at the temperature of a typical cell, which is 37 degrees Celsius. What happens to enzyme function when the temperature rises? What happens to enzyme function when the temperature drops?
Answer:
At high temperature the enzyme becomes denatured.
Lower temperature the enzymes become inactive.
Explanation:
Enzymes do not work at high temperature since the temperature kills the cells. And at LOW temperature enzymes have no energy to perform their work hence becomes inactive.
Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.20. If the molar absorptivity of yellow dye at 427 nm is 27400 M–1cm–1, what is the concentration of the solution?
Answer:
The concentration of the solution, [tex]C=7.2992\times 10^{-6} M[/tex]
Explanation:
The absorbance of a solution can be calculated by Beer-Lambert's law as:
[tex]A=\varepsilon Cl[/tex]
Where,
A is the absorbance of the solution
ɛ is the molar absorption coefficient ([tex]L.mol^{-1}.cm^{-1}[/tex])
C is the concentration ([tex]mol^{-1}.L^{-1}[/tex])
l is the path length of the cell in which sample is taken (cm)
Given,
A = 0.20
ɛ = 27400 [tex]M^{-1}.cm^{-1}[/tex]
l = 1 cm
Applying in the above formula for the calculation of concentration as:
[tex]A=\varepsilon Cl[/tex]
[tex]0.20= 27400\times C\times 1[/tex]
[tex]C = \frac{0.20}{27400\times 1} M[/tex]
So , concentration is:
[tex]C=7.2992\times 10^{-6} M[/tex]
The concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
BEER-LAMBERT EQUATION:
The concentration of a solution/sample measured using a spectrophotometer can be calculated using beer-lambert's equation as follows:A = εbc
Where;
ε = molar absorptivity of the yellow dye solution b = the path length of cuvettec = the concentration of the yellow dye solutionA = absorbance of the yellow dyec = A ÷ εb
According to this question, the absorbance of the yellow dye solution at 427 nm is 0.20, its molar absorptivity at 427 nm is 27400 M-¹cm-¹ and the cuvette length is 1.0cm. Hence, the concentration can be calculated as follows:c = A ÷ εb
c = 0.20 ÷ (27400 × 1)
c = 0.20 ÷ 27400
c = 7.29 × 10-⁶M
Therefore, the concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
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An initially evacuated 1.5 m tank is fed (adiabatically) with steam from a line available at a constant 15 MPa and 400 °C until the tank pressure reaches 15 MPa. What is the final mass of water in the tank in kg?
Answer:
95.8 kg
Explanation:
At the end of the feeding process, there is steam in the tank at 15 MPa and 400ºC because the process is adiabatic. So, use the steam tables (In this case I use data from van Wylen Six Edition, table B.13) in order to get the specific volume of superheated steam.
The specific volume data reported is [tex]v=0.01565\frac{m^{3}}{kg}[/tex]
The mass can be calculated with the definition of specific volume:
[tex]v=\frac{V}{m}\\m=\frac{V}{v}=\frac{1.5m^{3}}{0.01565\frac{m^{3}}{kg}} =95.8kg[/tex]
Calculate the pH of a 0.22 M ethylamine solution.
Answer:
answer is 12.18
Explanation:
(C2H5NH2, Kb = 5.6 x 10-4.)
Answer:
pH = 10.1
Explanation:
For weak base solutions [OH] = SqrRt([Base]·Kb
Then, from pH + pOH = 14 => pH = 14 - pOH
[OH} = SqrRt[(0.22)(5.6 x 10⁻⁴)] = 3.91
pH = 14 - 3.91 = 10.1
Be sure to answer all parts. The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is S(s) + O2(g) → SO2(g) If 2.68 × 107 tons of sulfur dioxide formed, how many tons of sulfur were present in the original materials? Assume 100% yield. × 10 tons Enter your answer in scientific notation.
Answer: The amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]
Explanation:
Converting given amount of mass in tons to grams, we use the conversion factor:
1 ton = 907185 g .......(1)
So, [tex]2.68\times 10^7=2.431\times 10^{13}g[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Given mass of sulfur dioxide = [tex]2.431\times 10^{13}g[/tex]
Molar mass of sulfur dioxide = 64 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of sulfur dioxide}=\frac{2.431\times 10^{13}g}{64g/mol}=3.79\times 10^9mol[/tex]
For the given chemical reaction:
[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of sulfur dioxide is produced from 1 mole of sulfur
So, [tex]3.79\times 10^9[/tex] moles of sulfur dioxide will be produced from = [tex]\frac{1}{1}\times 3.79\times 10^9=3.79\times 10^9[/tex] moles of sulfur.
Now, calculating the mass of sulfur using equation 2:
Moles of sulfur = [tex]3.79\times 10^9mol[/tex]
Molar mass of sulfur = 32 g/mol
Putting values in equation 2, we get:
[tex]3.79\times 10^9mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Moles of sulfur}=121.54\times 10^{11}g[/tex]
Converting this value in tons using conversion factor 1, we get:
[tex]\Rightarrow (\frac{1ton}{907185g})\times 121.54\times 10^{11}g\\\\\Rightarrow 13397491.6tons=1.34\times 10^7tons[/tex]
Hence, the amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]
Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X2O3 and write a balanced equation for the reaction.
Answer:The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Explanation:
Let the molecular formula of the oxide of metal be [tex]X_2O_y[/tex]
[tex]X_2O_y+yCO\rightarrrow yCO_2+2X[/tex]
Mass of metal product = 1.68 g
Moles of metal X =[tex]\frac{1.68 g}{55.9 g/mol}=0.03005 mol[/tex]
1 mol of metal oxide produces 2 moles of metal X.
Then 0.03005 moles of metal X will be produced by:
[tex]\frac{1}{2}\times 0.03005 mol=0.01502 mol[/tex] of metal oxide
Mass of 0.01502 mol of metal oxide = 2.40 g (given)
[tex]0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g[/tex]
y = 2.999 ≈ 3
The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Final answer:
To show the simplest formula of the oxide is X2O3, we calculate the moles of metal (X) and oxygen from given masses, find their ratio, and deduce the empirical formula. The balanced equation for the reaction with carbon monoxide is X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g).
Explanation:
To prove that the simplest formula of the oxide is X2O3, first we need to calculate the moles of metal X produced. Since the molar mass of X is given as 55.9 g/mol, we divide the mass of metal product (1.68 g) by the molar mass of X to obtain the number of moles:
moles of X = 1.68 g / 55.9 g/mol = 0.03005 mol
We know that the initial mass of the oxide is 2.40 g and the product (X) is 1.68 g, so the mass of oxygen in the oxide is:
mass of O = 2.40 g - 1.68 g = 0.72 g
rationalizing the ratio, we get approximately 2:3
Thus, the empirical formula of the oxide is X2O3.
Balanced Equation for the Reaction
The balanced equation for the reaction of metal X's oxide with carbon monoxide to obtain metal X and carbon dioxide is:
X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g)
This equation shows that the oxide of metal X reacts with carbon monoxide in a 1:3 mole ratio to produce the pure metal and carbon dioxide in a 2:3 mole ratio.
The weak base ammonia, NH3, and the strong acid hydrochloric acid react to form the salt ammonium chloride, NH4Cl. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.289 M solution of ammonium chloride at 25∘C
Final answer:
To calculate the pH of a 0.289 M ammonium chloride solution, first determine the Ka of the ammonium ion from the Kb of ammonia and set up an ICE table to find the hydronium ion concentration. Then calculate the pH using the negative logarithm of the hydronium ion concentration.
Explanation:
The pH of a solution of ammonium chloride can be calculated by first determining the Ka (acid dissociation constant) of the ammonium ion (NH4+), which is the conjugate acid of ammonia (NH3). Given that the Kb for ammonia is 1.8×10−5, the Ka for ammonium can be calculated using the relationship Ka = Kw/Kb, where Kw is the ion-product constant for water (1.0×10−14 at 25°C). In this case, Ka = 1.0×10−14 / 1.8×10−5 = 5.6×10−10.
Knowing the Ka, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentration of hydronium ions, H3O+, produced. Since ammonium chloride is a strong electrolyte, it completely dissociates in water, thus initial [NH4+] is 0.289 M, and initial [H3O+] is 0. After the equilibrium is established, we calculate the concentration of H3O+ and subsequently find the pH of the solution. The pH is determined using the formula pH = -log[H3O+]. For a solution of ammonium chloride, this results in an acidic pH due to the formation of H3O+ ions.
The pH of a 0.289 M solution of ammonium chloride is approximately 4.89
This is calculated by determining the dissociation constant (Ka) for NH₄⁺ and using the concentration of H₃O⁺ ions to find the pH.
The Ka is found using the relation to the base dissociation constant (Kb) of ammonia.To determine the pH of a 0.289 M solution of ammonium chloride (NH₄Cl), we need to consider the dissociation of the ammonium ion (NH₄⁺) in water:
NH₄⁺ (aq) + H₂O (l) ⇔ H₃O⁺ (aq) + NH₃ (aq)The equilibrium constant for this reaction is the acid dissociation constant (Ka) of the ammonium ion.
We can calculate Ka using the relation Ka = Kw / Kb.
Given Kw = 1.0 × 10⁻¹⁴ and Kb for ammonia (NH₃) as 1.8 × 10⁻⁵, we find:
Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰Assuming the dissociation of NH₄⁺ is small, the concentration of H₃O⁺ is 'x', and using the initial concentration of NH₄⁺ (0.289 M) in the expression for Ka:
Ka = [tex]\frac{[H_{3}O^{+}][NH_{3} ] }{NH_{4} ^{+} }[/tex]5.6 × 10⁻¹⁰ = [tex]\frac{(x)(x)}{(0.289 - x) }[/tex] ≈ [tex]\frac{x^{2} }{0.289}[/tex]Simplifying for 'x', we get:
x² = 5.6 × 10⁻¹⁰ * 0.289x² = 1.62 × 10⁻¹⁰x (which is [H₃O⁺]) = [tex]\sqrt{(1.62 \times 10^{-10} )}[/tex] ≈ 1.27 × 10⁻⁵ MThe pH is then calculated as:
pH = -log[H₃O⁺] ≈ -log(1.27 × 10⁻⁵) ≈ 4.89Hence, the pH of a 0.289 M solution of ammonium chloride is approximately 4.89