Answer: The amount of [tex]CO_2[/tex] emitted into the atmosphere is 0.808 moles.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For [tex]C_8H_{18}[/tex] :
Given mass of [tex]C_8H_{18}[/tex] = 11.6 g
Molar mass of [tex]C_8H_{18}[/tex] = 114.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }C_8H_{18}=\frac{11.6g}{114.23g/mol}=0.101mol[/tex]
The chemical equation for the combustion of octane follows:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
By stoichiometry of the reaction:
2 moles of octane produces 16 moles of carbon dioxide
So, 0.101 moles of octane will produce = [tex]\frac{16}{2}\times 0.101=0.808moles[/tex] of carbon dioxide.
Hence, the amount of [tex]CO_2[/tex] emitted into the atmosphere is 0.808 moles.
Write down the equation for the Lewis-Randall rule for fugacity of species in an ideal solution.
Answer :
Lewis-Randall rule : Lewis-Randall rule states that, the fugacity of a component is directly proportional to the mole fraction of the component in the solution in an ideal solution.
The equation used for the Lewis-Randall rule for fugacity of species in an ideal solution is :
[tex]f_i=X_i\times f_i^o[/tex]
where,
[tex]f_i[/tex] = fugacity in the solution
[tex]f_i^o[/tex] = fugacity of a pure component
[tex]X_1[/tex] = mole fraction of component
Final answer:
The Lewis-Randall rule indicates that the fugacity of a component in an ideal solution equals the product of the component's mole fraction and the fugacity of the pure component, central to thermodynamics and mixture behavior understanding.
Explanation:
The Lewis-Randall rule provides a foundation for understanding the fugacity of species in an ideal solution. It states that the fugacity of a component in an ideal mix is equal to the product of the mole fraction of that component in the mixture and the fugacity of the pure component at the same temperature and pressure. In mathematical terms, for a component i in an ideal solution, this can be expressed as fi = xi × fi°, where fi is the fugacity of the component in the mixture, xi is the mole fraction of the component, and fi° is the fugacity of the pure component.
This rule is pivotal in the field of thermodynamics and is particularly relevant when dealing with ideal solutions where the interactions between different species are similar to those present in pure substances. It provides a basis for calculating the fugacity coefficient, which is used to assess how the real behavior of a gas differs from the ideal predicted by Raoult's law in mixtures. The Lewis-Randall rule, alongside Raoult's and Henry's laws, forms a fundamental part of the theoretical framework for understanding phase equilibria and the behavior of mixtures.
You are given the following boiling point data. Which of the liquids would you expect to have the highest vapor pressure at room temperature?Ethanol, C2H5OH 78.5 °CEthylene glycol, C2H4(OH)2 198.0 °CDiethyl ether, C3H10O2 34.5 °CWater, H2O 100.0 °CMethanol, CH3OH 64.96 °C
Answer:
Diethyl ether
Explanation:
vapor pressure - the pressure exerted by the gaseous molecules on the walls of the container , is called its vapor pressure.
The compound with higher boiling point , will have lower vapor pressure ,
and the compound with lower boiling point , will have higher vapor pressure.
Hence, Boiling point and vapor pressure have inverse relation.
The vapor pressure and boiling point both, depends on the inter molecular interactions , i,e, the interaction between the molecules.
Since, the compound with stronger inter molecular interactions, will not easily convert to gas, hence will have higher boiling point between , therefore , its vapor pressure would be less.
But the compound with less inter molecular interactions , can easily vaporize to convert to gaseous state and hence will have lower boiling point, therefore, its vapor pressure would be higher .
Among all the options , diethyl ether have lowest boiling point , hence, will have highest vapor pressure , at room temperature.
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.
A solution is prepared by mixing equal volumes of 0.16 M HCl and 0.52 M HNO3. (Assume that volumes are additive.)
Express the pH to two decimal places.
Answer:
The pH of the final solution is 0.16 .
Explanation:
The pH of the solution is defined as negative logarithm of hydrogen ion concentration in a solution.
[tex]pH=-\log[H^+][/tex]
Concentration of HCl = 0.16 M
[tex]HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)[/tex]
HCl is a string acid .1 molar of HCl gives 1 molar of of hydrogen ions.
[tex][H^+]=0.16 M[/tex]
Concentration of [tex]HNO_3[/tex] = 0.52 M
[tex]HNO_3(aq)\rightarrow H^+(aq)+NO_{3}^-(aq)[/tex]
Nitric is a string acid .1 molar of nitric acid gives 1 molar of of hydrogen ions.
[tex][H^+]'=0.52 M[/tex]
Total hydrogen ion concentration:
[tex][H^+]''=[H^+]+[H^+]'[/tex]
=0.16 M+0.52 M=0.68 M
The pH of the solution:
[tex]\pH=-\log[H^+]''=-\log[0.68 M][/tex]
pH = 0.16
The pH of the final solution is 0.16 .
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
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.
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]
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.
Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.
Answer:
25.275 moles of oxygen gas will be required to completely react with all the samarium metal.
Explanation:
[tex]4Sm+3O_2\rightarrow 2Sm_2O_3[/tex]
Number of moles samarium metal = 33.7 moles
According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.
Then 33.7 moles of samarium will react with:
[tex]\frac{3}{4}\times 33.7 mol=25.275 mol[/tex]of oxygen gas.
25.275 moles of oxygen gas will be required to completely react with all the samarium metal.
Answer:
Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is [tex]\fbox{25.3 \text{ mol}}[/tex].
Explanation:
A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.
The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:
1. Amount of one reactant required to react completely with the other reactant.
2. Amount of the product that can be produced from the given amount of the reactant.
Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.
The chemical formula for oxygen gas is [tex]\text{O}_{2}[/tex].
Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is [tex]\text{Sm}_{2}\text{O}_{3}[/tex].
The chemical equation is as follows:
[tex]\fbox{\text{Sm}+\text{O}_{2} \rightarrow \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.
The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of [tex]\text{Sm}_{2}\text{O}_{3}[/tex] and 3 in front of [tex]\text{O}_{2}[/tex] to balance the oxygen atoms.
[tex]\fbox{\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.
[tex]\fbox{\\4\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]
Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.
According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.
Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:
[tex]\text{moles of O}_{2} = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)[/tex] ...... (1)
Step 5: Substitute 33.7 mol for moles of Sm in equation (1).
[tex]\text{Moles of O}_{2} = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)\\\text{Moles of O}_{2}= 25.275 \text{ mol}\\\text{Moles of O}_{2}= 25.3 \text{ mol}[/tex]
Note:
Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.
Learn more:
1. Balanced chemical equation https://brainly.com/question/1405182
2. Learn more about how to calculate moles of the base in given volume https://brainly.com/question/4283309
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Some basic concept of chemistry
Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.
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
The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 216.7 grams of benzene to reduce the vapor pressure to 71.61 mm Hg ? benzene = C6H6 = 78.12 g/mol.
Answer:
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
Explanation:
The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.
[tex]\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}[/tex]
Where:
[tex]p_o[/tex] = Vapor pressure of pure solvent
[tex]p_s[/tex] = Vapor pressure of the solution
[tex]n_1[/tex] = Number of moles of solvent
[tex]n_2[/tex] = Number of moles of solute
[tex]p_o = 73.03 mmHg[/tex]
[tex]p_s= 71.61 mmHg[/tex]
[tex]n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol[/tex]
[tex]\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}[/tex]
[tex]n_2=0.05499 mol[/tex]
Mass of 0.05499 moles of estrogen :
= 0.05499 mol × 272.4 g/mol = 14.9802 g
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
Final answer:
The answer involves using Raoult's law to calculate the mass of estradiol needed to lower the vapor pressure of benzene by a specified amount. The calculation uses the initial and final vapor pressures of benzene and the molecular weights of both compounds to solve for the mass of estradiol.
Explanation:The question relates to the calculation of the amount of a nonvolatile solute (estradiol) needed to be added to a solvent (benzene) to produce a desired lowering of the vapor pressure, via Raoult's law. Given the initial vapor pressure of benzene and the final vapor pressure after the addition of estradiol, we can use Raoult's law and the molecular weights of benzene and estradiol to solve for the mass of estradiol required.
To find the mass of estradiol that must be added to 216.7 grams of benzene to reduce its vapor pressure from 73.03 mm Hg to 71.61 mm Hg, we first calculate the mole fraction of benzene in the solution (XBenzene) after the addition of estradiol. Using the initial and final vapor pressures (Pinitial and Pfinal), and recognizing that the vapor pressure of the solution is directly proportional to the mole fraction of benzene, we get:
Pfinal = XBenzene × Pinitial
Then, we can calculate the moles of benzene (nBenzene) and use this value to find the moles of estradiol (nEstradiol) that correspond to the desired vapor pressure change. The mass of estradiol can then be computed by multiplying nEstradiol by its molecular weight (MW). Since estradiol is a nonvolatile nonelectrolyte, its addition does not alter the benzene vapor pressure other than through the change in mole fraction.
Calculate the number of milliliters of 0.472 M Ba(OH)2 required to precipitate all of the Ca2+ ions in 197 mL of 0.671 M CaBr2 solution as Ca(OH)2. The equation for the reaction is: CaBr2(aq) + Ba(OH)2(aq) Ca(OH)2(s) + BaBr2(aq)
Final answer:
To precipitate all of the Ca2+ ions in the CaBr2 solution, we need 0.280618 mL of 0.472 M Ba(OH)2.
Explanation:
To find the number of milliliters of 0.472 M Ba(OH)2 required to precipitate all of the Ca2+ ions, we first need to determine the number of moles of Ca2+ ions in the 197 mL of 0.671 M CaBr2 solution. Using the equation CaBr2(aq) + Ba(OH)2(aq) -> Ca(OH)2(s) + BaBr2(aq), we can see that the mole ratio between Ca2+ ions and Ba(OH)2 is 1:1. So if there are x moles of Ca2+ ions, we would need x moles of Ba(OH)2. Using the formula Molarity = Moles/Volume, we can calculate the moles of Ca2+ ions:
Moles of Ca2+ ions = (0.671 M CaBr2)(197 mL) = 0.132537 mol Ca2+
Since the mole ratio is 1:1, we would need 0.132537 mol of Ba(OH)2. Using the formula Moles = Molarity x Volume, we can calculate the volume of 0.472 M Ba(OH)2:
Volume of Ba(OH)2 = (0.132537 mol)/(0.472 M) = 0.280618 mL
What is the pH of 0.001 M H,SO4 (strong acid)? After mixing 250 ml 0.001 M H,SO, with 750 mL water, what is the pH now?
Answer : The pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.
Explanation :
First we have to calculate the concentration of hydrogen ion.
The balanced dissociation reaction will be,
[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex]
The concentration of [tex]H_2SO_4[/tex] = x = 0.001 M
The concentration of [tex]H^+[/tex] ion = 2x = 2 × 0.001 M = 0.002 M
The concentration of [tex]SO_4^{2-}[/tex] = x = 0.001 M
Now we have to calculate the pH of 0.001 M [tex]H_2SO_4[/tex].
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (0.002)[/tex]
[tex]pH=2.69[/tex]
Now we have to calculate the molarity after mixing the solution.
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = 0.001 M
[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 250 ml
[tex]M_2[/tex] = molarity of after mixing = ?
[tex]V_2[/tex] = volume of after mixing = 250 + 750 = 1000 ml
Now put all the given values in the above formula, we get the molarity after mixing the solution.
[tex](0.001M)\times 250ml=M_2\times (1000ml)[/tex]
[tex]M_2=2.5\times 10^{-4}M[/tex]
The concentration of [tex]H^+[/tex] ion = [tex]2\times (2.5\times 10^{-4}M)=5\times 10^{-4}M[/tex]
Now we have to calculate the pH after mixing the solution.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (5\times 10^{-4})[/tex]
[tex]pH=3.30[/tex]
Therefore, the pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.
Write the structure of the product that would be formed from the S(
S)-2-iodohexane and hydroxide ion. What would be formed if it underwent an Sr
eaction, with water as nucleophile?
Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.
The product formed on reaction with water would be a 50:50 mixture of
2S-hexane-2-ol. and 2R-hexane-2-ol.
Explanation:
2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good nucleophile .
The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.
In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.
When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .
The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.
The SN¹ reaction is a 2 step reaction , in the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.
In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.
The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.
Kindly refer the attachment for reaction mechanism and structure of products.
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.
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.
The reaction X 2 (g) m 2 X(g) occurs in a closed reaction vessel at constant volume and temperature. Initially, the vessel contains only X 2 at a pressure of 1.55 atm. After the reaction reaches equilibrium, the total pressure is 2.85 atm. What is the value of the equilibrium constant, Kp , for the reaction?
Final answer:
The value of the equilibrium constant, Kp, for the given reaction can be calculated based on the partial pressures of the reactants and products at equilibrium. In this case, the values are 1.425 atm for X₂ and 5.7 atm for X. Using the formula Kp = (p(X)²) / p(X₂), we find that Kp = 22.8.
Explanation:
The value of the equilibrium constant, Kp, for the reaction X₂(g) ⇌ 2X(g), can be calculated using the partial pressures of the reactants and products at equilibrium. In this case, the initial pressure of X₂ is 1.55 atm, and the total pressure at equilibrium is 2.85 atm.
Let's assume that the partial pressure of X₂ at equilibrium is p(X₂), and the partial pressure of X at equilibrium is p(X).
According to the equation, the number of moles of X₂ decreases by 1 (from 2 to 1) and the number of moles of X increases by 2 (from 0 to 2). This means that at equilibrium, the equilibrium partial pressure of X₂ is half of the total pressure, and the equilibrium partial pressure of X is twice the total pressure.
Therefore, p(X₂) = 2.85/2 = 1.425 atm and p(X) = 2 * 2.85 = 5.7 atm.
To calculate the value of Kp, we use the formula:
Kp = (p(X)²) / p(X₂)
Substituting the values we obtained, we get:
Kp = (5.7²) / 1.425 = 22.8
In the following pairs of compounds, which is the most acidic? Benzoic acid and 4-nitrobenzoic acid 4-Methylbenzoic acid and 4-chlorobenzoic acid p-Nitrophenol and m-nitrophenol
Answer:4-nitrobenzoic acid is more acidic than Benzoic acid
4-chlorobenzoic acid is more acididc than 4-methyl benzoic acid
P-Nitrophenol is more acidic than meta-nitrophenol
Explanation:Acidity of an acid can be explained in terms of the stability of conjugate base formed.
1. 4-nitrobenzoic acid is more acidic as compared to benzoic acid because of the presence of nitro group at 4-position that is para position of the benzene ring. Nitro group is an electron withdrawing group and it withdraws the electron density through resonance effect.
Here the conjugate base would be benzoate anion which has a carboxylate anion attached with the benzene ring . So any group which can withdraw the electron density from benzoate anion will stabilise the benzoate anion and subsequently it would increase the acidity .
In case of benzoic acid there is no extra withdrawl of electron density whereas in case of 4-nitrobenzoic acid the nitro group stabilises the benzoate anion by withdrawing electron density thereby stabilising the benzoate anion and increasing the acidity.
2. 4-chlorobenzoic acid is more acidic than 4-methyl benzoic acid because 4-chlorobenzoic acid has Cl group which is a good electron withdrawing group through inductive effect so the benzoate anion formed can be stabilised by the electron withdrawing Cl atom which would increase the acidity of 4-chlorobenzoic acid.
4-methylbenzoic acid has an electron donating Methyl group which donates electron density through inductive effect hence a methyl group would intensify the negative charge on the benzoate anion through electron donation and subsequently it would destabilise the benzoate anion thereby decreasing its acidity.
3. In case of phenols the conjugate base formed is phenoxide anion and the negative charge that is its electron density is delocalised over the whole phenol ring. The negative charge electron density is more prominent at ortho and para position rather than the meta position. p-nitrophenol is more acidic than m-nitrophenol because p-nitrophenol has the nitro group at para position where it can stabilise the phenoxide anion through electron withdrawl via resonance whereas in case of m-nitrophenol as the nitro group is present at meta position so it can not stabilize prominently through electron withdrawl via resonance.
In each pair, 4-nitrobenzoic acid, 4-chlorobenzoic acid, and p-nitrophenol are more acidic due to their electron withdrawing groups that can better stabilize the negative charge after ionization.
Explanation:The acidity of a compound can be determined by its ability to donate a proton (hydrogen ion), and this ability is often influenced by other groups present in the compound. In each pair, the compound with the group that is more electron withdrawing will generally be more acidic because they can better stabilize the negative charge of the carboxylate ion after ionization occurs.
Benzoic acid vs. 4-nitrobenzoic acid: The presence of the nitro group in the 4-nitrobenzoic acid makes it more electron withdrawing compared to benzoic acid, making it more acidic.
4-Methylbenzoic acid vs. 4-chlorobenzoic acid: The chlorine atom in 4-chlorobenzoic acid is more electron withdrawing than the methyl group in 4-methylbenzoic acid, making 4-chlorobenzoic acid more acidic.
p-Nitrophenol vs. m-nitrophenol: In this case, p-nitrophenol would be more acidic. Although the nitro group is meta to the phenol group in m-nitrophenol, and para in p-nitrophenol, the para position allows for more effective resonance stabilization post-ionization, enhancing its acidity.
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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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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.
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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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.
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.
A steady-state plasma concentration of 25 mg/L v by IV infusion to healthy volunteers (average weight, 75 kg) at a rate of 7.5 mg/kg/hr for 6 hours. Calculate the total body clearance of this drug. A. 12.5 L/hr B. 22.5 L/hr C. 42.5 L/hr D. 62.5 L/hr was measured when a drug was given
Answer:
The correct answer is option B.
Explanation:
Concentration of a steady-state plasma ,= 25 mg/L
Weight of the volunteer =75 kg
Rate of infusion = 7.5 mg/kg hr
Concentration of steady-state plasma in 75 kg weight body:
[tex]7.5 mg/kg hr\times 75 kg=562.5 mg/hr[/tex]
Total body clearance of this drug:
[tex]\frac{562.5 mg/hr}{25 mg/L}=22.5 L/hr[/tex]
The total body clearance of this drug is 22.5 L/hr.
A scientist measures the standard enthalpy change for the following reaction to be 595.8 kJ : 2H2O(l)2H2(g) + O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is kJ/mol.
Answer: The [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.
The equation that is used to calculate enthalpy change of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(H_2)})+(1\times \Delta H_f_{(O_2)})]-[(2\times \Delta H_f_{(H_2O)})][/tex]
We are given:
Enthalpy of substances present in their standard form are taken to be 0.
[tex]\Delta H_f_{(H_2)}=0kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=595.8kJ[/tex]
Putting values in above equation, we get:
[tex]595.8=[(2\times (0))+(1\times (0))]-[2\times (\Delta H_f_{H_2O})]\\\\\Delta H_f_{H_2O}=297.9kJ/mol[/tex]
Hence, the [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.
In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to be 5.153. Use the information she obtained to determine the Ka for this acid.
Answer:
The dissociation constant of phenol from given information is [tex]9.34\times 10^{-11}[/tex].
Explanation:
The measured pH of the solution = 5.153
[tex]C_6H_5OH\rightarrow C_6H_5O^-+H^+[/tex]
Initially c
At eq'm c-x x x
The expression of dissociation constant is given as:
[tex]K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}[/tex]
Concentration of phenoxide ions and hydrogen ions are equal to x.
[tex]pH=-\log[x][/tex]
[tex]5.153=-\log[x][/tex]
[tex]x=7.03\times 10^{-6} M[/tex]
[tex]K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}[/tex]
[tex]K_a=9.34\times 10^{-11}[/tex]
The dissociation constant of phenol from given information is [tex]9.34\times 10^{-11}[/tex].
Final answer:
The acid dissociation constant (Ka) for phenol, given its pH of 5.153 in a 0.529 M solution, is calculated to be 9.47 x 10⁻¹² M by converting pH to hydrogen ion concentration and applying the equilibrium expression for dissociation.
Explanation:
To determine the acid dissociation constant (Ka) for phenol from its pH, we use the formula pH = -log[H⁺], where [H⁺] is the concentration of hydrogen ions in the solution. First, convert the pH given (5.153) to [H+] concentration: [H⁺] = 10^-pH = 10^-5.153 = 7.08 x 10⁻⁶ M. The ionization of phenol in water can be represented as: C₆H₅OH (aq) → C₆H₅O⁻ (aq) + H⁺ (aq). At equilibrium, the concentration of C6H5O- and H+ will be equal and be 7.08 x 10-6 M, as the weak acid only partially dissociates. The initial concentration of phenol is 0.529 M, so the concentration of undissociated phenol at equilibrium is approximately 0.529 M (assuming the dissociation is minimal due to the weak acid nature). The Ka for phenol can be calculated using the expression Ka = [H⁺][C₆H₅O⁻]/[C₆H₅OH], substituting the equilibrium concentrations yields Ka = (7.08 x 10⁻⁶)2/0.529 = 9.47 x 10⁻¹² M.
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]
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.
Calculate the average molecular weight of air (1) from its approximate molar composition of 79% N2, 21% 02x
Answer:
Average molecular weight of air is 28.84 g/mol.
Explanation:
The average molecular weight of a mixture is determined from their molar composition and molecular weight.
Average molecular weight :[tex]\sum (\chi_i\times m_i)[/tex]
[tex]\chi_1[/tex] : mole fraction of the 'i' component.
[tex]m_i[/tex] = Molecular weight of i component
Average molecular weight of air with approximate molar composition of 79% nitrogen gas and 21% of oxygen gas can be calculated as:
Average molecular weight of air:
[tex]79\%\times 28 g/mol+21\%\times 32 g/mol[/tex]
[tex]=0.79\times 28 g/mol+0.21\times 32 g/mol=28.84 g/mol[/tex]
Final answer:
The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2.
Explanation:
The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2. The molar weight of N2 is 28.01 g/mol and the molar weight of O2 is 32.00 g/mol. We can calculate the average molecular weight using the formula:
Average Molecular Weight = (0.79 * 28.01 g/mol) + (0.21 * 32.00 g/mol)
Average Molecular Weight = 28.89 g/mol
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.