Answer:
A. pH using molar concentrations = 2.56
B. pH using activities = 2.46
C. pH of mixture = 2.56
Explanation:
A. pH using molar concentrations
ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺
HA + H₂O ⇌ A⁻ + H₃O⁺
We have a solution of 0.08 mol HA and 0.04 mol A⁻
We can use the Henderson-Hasselbalch equation to calculate the pH.
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}[/tex]
B. pH using activities
(i) Calculate [H⁺]
pH = -log[H⁺]
[tex]\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}[/tex]
(ii) Calculate the ionic strength of the solution
We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.
The formula for ionic strength is
[tex]I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041[/tex]
(iii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79[/tex]
(iv) Calculate the initial activity of A⁻
a = γc = 0.79 × 0.04= 0.032
(v) Calculate the pH
[tex]\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\[/tex]
C. Calculate the pH of the mixture
The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.
The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.
The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.
The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.
(i) Calculate the ionic strength
[tex]I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10[/tex]
(ii) Calculate the activity coefficients
[tex]\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69[/tex]
(iii) Calculate the initial activity of A⁻:
a = γc = 0.69 × 0.05= 0.034
(iv) Calculate the pH
[tex]\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}[/tex]
what volume in ML of 12.0M HCL is needed to contain 3.00 moles of HCL?
Answer:
V= 250ml
Explanation:
From n= CV
3= 12×V
V= 0.25L= 250ml
1. What values are needed to determine the energy of an electron in a many‑electron atom?
O n
O ????
O m????
O ms
2. What information is most important in determining the size of an orbital?
O n
O ????
O m????
O ms
3. What information is needed to determine the orientation of an orbital?
O n
O ????
O m????
O ms
4. What information is needed to determine the general shape of an orbital?
O n
O ????
O m????
O ms
Answer: 1. n
2. n
3. [tex]m_s[/tex]
4. l
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1).For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_s[/tex] . The value of this quantum number ranges from -l to +l.
Spin Quantum number : It describes the direction of electron spin. This is represented as s.
To determine the energy of an electron in a many-electron atom, the principal and angular momentum quantum numbers are needed (n and l). The principal quantum number (n) primarily determines the size of an orbital, while the magnetic quantum number (ml) is needed for its orientation. The angular momentum quantum number (l) is responsible for the general shape of an orbital.
To determine the energy of an electron in a many-electron atom, the following quantum numbers are needed:
n (principal quantum number) determines the general range for the value of energy and the probable distances that the electron can be from the nucleus.I (angular momentum quantum number) helps to determine subshell energy levels.The most important information in determining the size of an orbital is:
n (principal quantum number), which largely determines the energy and size of the orbital.To determine the orientation of an orbital, the quantum number needed is:
ml (magnetic quantum number), which describes the orientation of the orbital in space.The quantum number needed to determine the general shape of an orbital is: I (angular momentum quantum number), which describes the shape or type of the orbital.
Each electron also has a spin quantum number, ms, which determines the spin of an electron and can have a value of either +1/2 or -1/2.
A student placed 11.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution
Answer:
There is 0.92 g of glucose in 100 mL of the final solution.
Explanation:
Initially, 11.5 g of glucose is added to the volumetric flask
Water is then added to 100 mL Mark,
The flask was then shaken until the solution was uniform.
The shaking of the mixture makes the concentration of glucose to become uniform all through the solution.
At this point, the concentration of this solution in g/mL is (11.5/100) = 0.115 g/mL
A 40.0 mL sample of this glucose solution was diluted to 0.500 L.
40.0 mL of the already mixed solution is then diluted to 0.500 L.
The mass of glucose in 40.0 mL of the mixed solution with concentration 0.115 g/mL is then given as
Mass = (conc in g/mL) × (volume) = 0.115 × 40 = 4.6 g
So, this mass is then diluted to 0.500 L mark.
New concentration = (mass)/(conc In mL) = (4.6/500) = 0.0092 g/mL
How many grams of glucose are in 100. mL of the final solution
Mass = (conc in g/mL) × (Volume in mL) = 0.0092 × 100 = 0.92 g
Hope this Helps!!!
Answer:
0.459 gram
Explanation:
Find the attachment
Which statement best describes what is taking place? Copper is being oxidized. Copper is being reduced. Copper is losing electrons. Copper is a reducing agent.
Answer:
copper is being reduced
Explanation: I got 100% on the quiz good luck this class is hard lol
Answer:
The answer is the second option
Copper is being reduced
Explanation:
Have a good day just took test
A mixture of krypton and neon gas is compressed from a volume of 96.0L to a volume of 68.0L , while the pressure is held constant at 15.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.
Answer:
-42556.5 J
Explanation:
From gas law,
The work done by gas is given as,
W = PΔV......................... Equation 1
Where W = Work done on the gas mixture, P = Pressure of the gas mixture, ΔV = Change in volume of the gas mixture.
Given: P = 15 atm = (15×101325) N/m² = 1519875 N/m², ΔV = 68-96 = -28 L = -28/1000 = -0.028 m³
Substitute into equation 2
W = 1519875(-0.028)
W = -42556.5 J
Hence the work done by the gas mixture = -42556.5 J
The work performed on the gas mixture of krypton and neon when compressed at a constant pressure of 15.0 atm from a volume of 96.0L to 68.0L is -4.26 x 10⁴ joules.
Explanation:The subject question asks to compute the work done on a gas mixture comprising krypton and neon as it's compressed from a volume of 96.0L to 68.0L, with a constant pressure of 15.0 atm. The calculation of work done on gas under constant pressure relies on the formula W = -P ΔV, where W is work, P represents pressure, and ΔV is the change in volume. In this problem, P = 15.0 atm, which we will convert to pascals (1 atm = 1.01 × 10⁵ Pa). The change in volume ΔV = Vf - Vi = 68.0L - 96.0L = -28.0L, which we will convert to cubic meters (1L = 1.0 × 10-³ m³). Applying these values, W = -P ΔV = -(15.0atm) * (-28.0L) = 420 L*atm. To convert this to joules, we can use the conversion factor 101.3 J = 1 L*atm, which results in a final answer of -4.26 x 10⁴ J. The result is negative as it indicates the work is done on the gases.
Learn more about Work done on gas here:https://brainly.com/question/32263955
#SPJ3
A sample of gas contains 0.1100 mol of N2(g) and 0.3300 mol of H2(g) and occupies a volume of 20.5 L. The following reaction takes place: N2(g) 3 H2(g) 2 NH3(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. g
Final answer:
To find the volume of the sample after the given reaction, Avogadro's law is used to show that the volume of gases at constant temperature and pressure is directly proportional to the number of moles. The balanced equation indicates that two times the volume of NH3 gas is produced from one volume of N2 gas. The final volume of the sample is calculated to be 10.25 L after the reaction.
Explanation:
To determine the volume of the sample after the reaction between N2 and H2, we must first understand the stoichiometry of the reaction. The balanced chemical equation for the synthesis of ammonia is:
N2(g) + 3H2(g) → 2NH3(g)
According to Avogadro's law, at constant temperature and pressure, equal volumes of gases contain an equal number of molecules. Thus, in our reaction, one volume of N2 reacts with three volumes of H2 to produce two volumes of NH3. In this case, we begin with 0.1100 mol of N2 and 0.3300 mol of H2. Since H2 is present in excess (needed only 3 times the amount of N2), all of the N2 will be consumed first.
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
Final answer:
By applying Avogadro's law, we determine that all of the initial N2 and H2 reacts to form NH3. Because equal volumes of any gas at the same temperature and pressure have the same number of molecules, we understand that the volume of NH3 produced will be equivalent to the volume of N2 reactant consumed plus three times this volume due to 3H2 reacting. Therefore, the final volume should be less than the initial volume since 2 moles of NH3 will be produced from 4 moles of combined reactants.
Explanation:
Understanding Reaction Volumes and Avogadro's Law:
Based on the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can apply Avogadro's law to determine the change in volume when the reaction occurs at constant temperature and pressure.
At the onset, we have 0.1100 mol of N2(g) and 0.3300 mol of H2(g) occupying a combined volume of 20.5 L. Assuming the reaction goes to completion, we start by identifying the limiting reactant, the reactant that will be completely consumed in the reaction. The stoichiometry of the reaction indicates that for every 1 mol of N2, 3 mol of H2 are required.
For 0.1100 mol of N2, we would need 0.3300 mol of H2 to fully react based on the stoichiometry of 1:3. Since we have exactly 0.3300 mol of H2, it means N2 is the limiting reactant and all of it will be converted into NH3.
Now, according to Avogadro's law, if one mole of N2 gas occupies a certain volume and reacts to form two moles of NH3 at the same conditions of temperature and pressure, the volume of NH3 produced will double that of the N2 because two moles of NH3 contain double the number of molecules compared to one mole of N2. However, since H2 also takes up volume and we have three moles of H2 reacting for every mole of N2, the total volume initially is the volume of N2 plus three times the volume of N2 (which is the volume of H2).
The stoichiometry tells us that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. So 0.1100 mol of N2 would produce 0.2200 mol of NH3. Since we are assuming the temperature and pressure are constant, we can use the ratio of moles to determine the change in volume. Initially, there are 0.1100 + 0.3300 = 0.4400 mol of gases and after reaction we have 0.2200 mol of NH3.
The initial volume is 20.5 L. With the moles reducing from 0.4400 to 0.2200, the volume occupied by the gases after the reaction is expected to be half of the initial volume, given the relationship V1/n1 = V2/n2 where V is volume and n is the number of moles. Therefore, the volume of the sample after the reaction is 10.25 L.
How many moles of NH3 are required to produce 12 moles of NH4Cl
Answer:
16 moles
Explanation:
16 moles of NH₃ are required to produce 12 moles of NH₄Cl
What is Stoichiometry ?Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.
What is Balanced Chemical Equation ?The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
First we have to write the balanced chemical equation.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
Here 8 moles of NH₃ produce 3 moles of Cl₂ to form 6 moles of NH₄Cl.
So,
12 mole NH₄Cl × [tex]\frac{8\ \text{mole}\ NH_3}{6\ \text{mole}\ NH_4Cl}[/tex]
= 16 moles NH₃
Thus, we can say that 16 moles of NH₃ are required to produce 12 moles of NH₄Cl.
Learn more about the Stoichiometry here: brainly.com/question/14935523
#SPJ2
Aldol condensation of 2,5-heptanedione yields a mixture of two enone products in a 9:1 ratio. Treatment of the minor product with aqueous NaOH converts it into the major product; the interconversion proceeds as follows: Hydroxide ion adds to the double bond, forming enolate ion 1; Proton transfer occurs, yielding tetrahedral intermediate 2; Ring opening occurs, yielding enolate ion 3; Protonation of enolate ion 3 occurs, yielding 2,5-heptanedione; Deprotonation at C-6 occurs, yielding enolate ion 5; Enolate ion 5 attacks C-2, yielding tetrahedral intermediate 6; Protonation occurs to yield aldol addition product 7; Dehydration yields the more stable product.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The diagram of the mechanism of this reaction is shown on the second uploaded image
The structure of the enolate Ion 1 is shown on the third uploaded image
Explanation:
The aldol condensation of 2,5-heptanedione involves several steps of proton transfer and structural changes to yield stable products. These steps showcase the dynamic nature of chemical reactions and examples of sequential proton transfers seen in polyprotic acids.
Explanation:The aldol condensation of 2,5-heptanedione yielding a mixture of enone products is a chain of reactions involving the transfer of protons. Iterating the process you described, it starts with hydroxide ion adding to the double bond to form intermediate enolate ion 1, followed by proton transfer resulting in tetrahedral intermediate 2. The ring opening process then forms another enolate ion (3), which after protonation, yields the initial 2,5-heptanedione. A deprotonation at C-6 subsequently yields enolate ion 5, which attacks C-2 to generate tetrahedral intermediate 6. Protonation at this stage will yield aldol addition product 7. Dehydration of this product will finally yield the more stable product. This sequence is an example of sequential proton transfers often seen in polyprotic acids.
Learn more about Aldol Condensation here:https://brainly.com/question/31433533
#SPJ3
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the ratio of the ionic radius of the cation to the ionic radius of the anion is 0.840, what is the ionic radius of the anion
Answer:
the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
Explanation:
From the diagram shown below :
The anion [tex]Cl^-[/tex] is located at the corners
The cation [tex]Cs^+[/tex] is located at the body center
The Body diagonal length = [tex]\sqrt{3 \ a }[/tex]
∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]
Given that :
[tex]\frac{r^+}{r^-} =0.84[/tex] (i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )
[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]
Also ; a = 664 pm
Then :
[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]
Therefore, the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
The ionic radius of the anion [tex]r^-=312.52pm[/tex]
Primitive cubic structure:The anion is placed on the corners and the cation is placed on the frame center.
The Body diagonal length = [tex]\sqrt{3a}[/tex]
[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]
Given:
Ratio= 0.840
[tex]\frac{r^+}{r^-}=0.840[/tex]
[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]
Also ; a = 664 pm
Then : [tex]r^-[/tex] =312.52 pm
Therefore, the ionic radius of the anion = 312.52 pm
Find more information about Cubic structure here:
brainly.com/question/7959646
A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of ethanolamine is ·1.02gcm−3. Calculate the volume of ethanolamine the student should pour out.
To determine the volume of ethanolamine needed, the student should divide the desired mass (10.0g) by the density of ethanolamine (1.02g/cm³), which gives approximately 9.8 cm³. A trusted source, such as the CRC Handbook of Physics and Chemistry was referenced.
Explanation:To calculate the volume of ethanolamine needed for the experiment, we need to use the formula for density, which is mass/volume. By rearranging the formula to solve for volume gives us volume = mass/density. So, the volume of ethanolamine will be 10.0g / 1.02g/cm3. This calculation gives us approximately 9.80 cm3. Therefore, the student needs to pour out about 9.8 cm3 of ethanolamine for the experiment. The student was able to easily determine this by consulting a trusted resource, the CRC Handbook of Physics and Chemistry.
Learn more about Volume Calculation here:https://brainly.com/question/32822827
#SPJ3
Final answer:
To find the volume of ethanolamine needed, divide the mass needed (10.0 g) by the density (1.02 g/cm³) to obtain approximately 9.80 cm³.
Explanation:
To calculate the volume of ethanolamine the student should pour out, we can use the density formula, which is density (d) equals mass (m) divided by volume (V), rearranged to solve for volume. Given that the density of ethanolamine is 1.02 g/cm³ and the student needs 10.0 g of ethanolamine, the equation would be:
V = m / d
Substituting the given values, we get:
V = 10.0 g / 1.02 g/cm³
The calculation results in a volume of approximately 9.80 cm³ of ethanolamine the student should pour out.
A 60.0 g aluminum block, initially at 55.00 °C, is submerged into an unknown mass of water at 293.15 K in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 25.00 °C. What is the approximate mass of the water? The specific heat of water is 4.18 J/g . °C. The specific heat of aluminum is 0.897 J/g . °C.
Answer:
The approximate mass of the water is 80kg
Explanation: Heat lost=heat gained
M1c1(Ʃ)=M2c2(Ʃ)
M1 is mass of aluminum
M2 is the mass of water
C1 is specific heat capacity of aluminum
C2 is specific heat capacity of water
Ʃ is change in temperature.
60 x0.897 x(55-25)=M2 x 4.18 x (25-20.15)
1614.6=20.27M2
M2=79.65
M2=80kg
t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
Complete Question:
t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be
I + S ⇄ I*S
W + S ⇄ W*S
W*S + I*S ⇄ TBA * S * S
TBA * S ⇄ TBA + S
Derive a rate law assuming:
(a) The surface reaction is rate limiting
(b) The adsorption of isobutene is limiting
(c) The reaction follows Eley-rideal Kinetics
I*S+W ⇄ TBA * S
and surface reaction is limiting
(d) Isobutene (I) and water (W) are adsorbed on different sites
I + S₁ ⇄ I*S₁
W + S₂ ⇄ W*S₂
TBA is not on the surface, and the surface reaction is rate-limiting
[tex][Ans: r'_{TBA}=-r'_1=\frac{k[C_1C_w-C_{TBA/K_C}]}{(1+K_WC_W)(1+K_1C_1)} ][/tex]
(e) What generalizations can you make by comparing rate laws derived from part (a) through (d)?
Answer and explanation:
The mechanism for the production of t-butyl alcohol is as follows:
the reaction and rate law for the adsorption of isobutene over the amberlyst-15 is as follows:
I + S ⇄ I * S [tex]-r_{ADI} = k_I(C_1C_v-\frac{C_{I.S}}{K_I} )[/tex]
where [tex]C_V[/tex] is the concentration of vacant site
[tex]K_I[/tex] is the equilibrium constant of the adsorption
[tex]k_I[/tex] is the rate constant for forward
[tex]C_I,C_{I.S}[/tex] are concentration of isobutene and site filled with isobutene
The reaction and rate law for the adsorption of water (W) over the amberlyst-15 catalyst catalyst is as follows
W + S ⇄ W.S [tex]-r_{ADW} = k_W(C_WC_V-\frac{C_{W.S}}{K_W} )[/tex]
The reaction and rate law for the surface reaction on the catalyst is as follows
W.S + I.S ⇄ TBA . S + Sn [tex]-r_s = k_s(C_{W.S}C_{I.S}-\frac{C_{TBA.S}C_V}{K_s} )[/tex]
The reaction and rate law for the desorption of TBA from catalyst is as follows
TBA . S ⇄ TBA + S [tex]-r_{D TBA} = k_{DTBA}(C_{TBA.S}-\frac{C_{TBA}C_V}{K_{DTBA}} )[/tex]
the attached image below gives the remaining steps
TBA, an octane enhancer, is produced by the hydration of isobutene over an Amberlyst-15 catalyst, resulting in a multiphase mixture that can be made miscible with the help of cosolvents or excess TBA.
Explanation:The production of t-Butyl Alcohol (TBA) involves a liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The question provides background information on the reaction mechanism and mentions the use of cosolvents or excess TBA to achieve reasonable miscibility in the multiphase mixture. This process results in a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can aid in achieving miscibility. Essentially, the production of TBA serves as a safer alternative to lead additives in gasoline, enhancing octane levels.
Learn more about t-Butyl Alcohol production here:https://brainly.com/question/36895915
#SPJ3
The hydroxyl radical, a fragment of water vapor known as the "atmospheric cleanser", with a chemical formula of OH, has a globally averaged number density of ~1×10^6 molecules cm-3 that is fairly constant with altitude. Determine the OH mixing ratio (in ppt) at the surface, where total pressure P ~1 atm, temperature T ~282K, and also at 10 km altitude, where P ~0.260 atm,
Answer:
Check the explanation
Explanation:
The hydroxyl radical, •OH, is the hydroxide ion (OH−) when in neutral form of the Hydroxyl radicals they are extremely reactive (easily becoming hydroxy groups) and as a result are short-lived; however, they form a significant part of radical chemistry. The hydroxyl radical composition is also highly reactive towards oxidative reactions.
Kindly check the attached image below to see the step by step explanation to the question above.
Given that the rate of diffusion of nickel in iron is very much greater in the liquid state than in the solid state, what effect should this have on the ease of obtaining an equilibrium microstructure (i.e., one that is homogeneous) when an alloy containing the peritectic composition 4.5 percent nickel is cooled through the peritectic temperature
Answer:
Explanation:
The rate of diffusion of nickle in is higher in liquid state than solid state which is affect the ease of equilibrium of microstructure.
When peritectic composition of 4.5% nickle is called from peritectic temperature due to high rate of diffusion of nickle into iron we get fine microstructure containing more nickle atoms in the iron.
Due to high rate of diffusion more no. Of microstructure created from where new grain is generated. So the microstructure will get equilibrium soon after cooling.
Answer:
Explanation:
The rate of diffusion of nickle in is higher in liquid state than solid state which is affect the ease of equilibrium of micro structure.
When peritectic composition of 4.5% nickle is called from peritectic temperature due to high rate of diffusion of nickle into iron we get fine micro structure containing more nickle atoms in the iron.
Due to high rate of diffusion more no. Of micro structure created from where new grain is generated. So the micro structure will get equilibrium soon after cooling.
How many moles of nitrogen are there in a 16,500 mL sample of nitrogen at STP?
Final answer:
To find the number of moles of nitrogen in a 16,500 mL sample at STP, convert the volume to liters and divide by the molar volume of 22.4 L/mol, resulting in approximately 0.7366 moles of nitrogen.
Explanation:
To calculate the number of moles of nitrogen in a 16,500 mL sample of nitrogen at STP (Standard Temperature and Pressure), we should use the molar volume of a gas at STP, which is 22.4 L/mol. This means that at STP, 1 mole of any gas occupies 22.4 liters.
First, we need to convert the volume of nitrogen from milliliters to liters:
16,500 mL × (1 L / 1,000 mL) = 16.5 L
Then, we divide the volume of nitrogen by the molar volume:
16.5 L / 22.4 L/mol = 0.7366 moles
Therefore, the sample contains approximately 0.7366 moles of nitrogen gas.
1. Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below. 1) 1-bromohexane 2) 3-bromo-3-methylpentane 3) 1-bromo-2,2-dimethylbutane 4) 3-bromo-2-methylpentane 5) 2-bromo-3-methylpentane a) the compound(s) that can exist as enantiomers b) the compound(s) that can exist as diastereomers c) the compound that gives the fastest SN2 reaction with sodium methoxide d) the compound that is least reactive to sodium methoxide in methanol e) the compound(s) that undergo an SN1 reaction to give rearranged products f) the compound that gives the fastest SN1 reaction
Answer:
See explanation
Explanation:
the compound that gives the fastest SN2 reaction with sodium methoxide- 1-bromohexane
the compound that gives the fastest SN1 reaction- 3-bromo-3-methylpentane
the compound(s) that undergo an SN1 reaction to give rearranged products- 1-bromo-2,2-dimethylbutane
the compound that is least reactive to sodium methoxide in methanol -
3-bromo-3-methylpentane
the compound(s) that can exist as diastereomers - 3-bromo-3-methylpentane
the compound(s) that can exist as enantiomers- 3-bromo-2-methylpentane
At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of thiophene and of acetyl bromide . Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. However the methodology will remain the same.
First we calculate the moles of thiophene and heptane, using their molar mass:
137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene 111 g heptane ÷ 100 g/mol = 1.11 moles heptaneTotal number of moles = 1.63 + 1.11 = 2.74 moles
The mole fraction of thiophene is:
1.63 / 2.74 = 0.59Finally, the partial pressure of thiophene vapor is:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
Partial Pressure = 0.59 * 0.60 atmPp = 0.35 atmFinal answer:
The partial pressure of thiophene vapor above an ideal solution can be determined by calculating the mole fraction of thiophene and then applying Raoult's Law, which involves multiplying the mole fraction by the vapor pressure of pure thiophene.
Explanation:
To calculate the partial pressure of thiophene vapor above the solution, we can utilize Raoult's Law, which pertains to ideal solutions. According to Raoult's Law, the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution (mole fraction). To find the mole fraction of thiophene, we will sum up the total moles of both thiophene and acetyl bromide and use their respective amounts to calculate the mole fractions for each.
First, we calculate the mole fraction of thiophene (Xthiophene) in the solution:
Xthiophene = moles of thiophene / (moles of thiophene + moles of acetyl bromide)Then, we apply Raoult's Law to find the partial pressure of thiophene vapor (Pthiophene):Pthiophene = Xthiophene × vapor pressure of pure thiopheneThe partial pressure of thiophene vapor can therefore be obtained, ensuring we keep consistent units and consider significant digits as per the given values in the question.
Your teacher needs a 3.0M solution acid, but only has a 12.0M stock solution of sulfuric acid in the chemical store room. Calculate and describe the steps the teacher needs to take in order to make 100mL of the 3.0M solution of sulfuric acid.
Answer:
The calculations are in the explanation below.
The steps are:
1. Using a graduated pipette, accurately take 25mL of the 12.0M stock solution.2. Pour the 25mL of stock solution into a 100 mL volumetric flask3. Add distilled water up to the mark4. Cap the flask with the stopper5. Stirr by gently rotating the flask.Explanation:
To make 100 mililiter of the 3.0M solution of sulfuric acid, first you must calculate the volume of the 12.0M stock solution that contains the same number of moles as the diluted solution.
For that, you use the dilution formula:
number of moles = C₁V₁ = C₂V212.0M×V₁ = 3.0M × 100mLV₁ = 3.0M × 100mL/12.0M = 25mLThen, the steps are:
1. Using a graduated pipette, accurately take 25mL of the 12.0M stock solution.
2. Pour the 25mL of stock solution into a 100 mL volumetric flask
3. Add distilled water up to the mark
4. Cap the flask with the stopper
5. Stirr by gently rotating the flask.
a gas at 80kPa occupies a volume of 5mL. WHat volume will the gas occupy at 70kPa?
Answer:
Volume = 5.71mL
Explanation:
Applying Boyle's law
P1V1= P2V2
P1= 80kPa, V1= 5mL,P2= 70kPa, V2=?
Substitute into above formula
80×5= 70×V2
V2= (80×5)/70 = 5.71mL
When the epoxide 2-vinyloxirane reacts with lithium dibutylcuprate, followed by protonolysis, a compound A is the major product formed. Oxidation of A with PCC yields B, a compound that gives a positive Tollens test and has an intense UV absorption around 215 nm. Treatment of B with Ag2O, followed by catalytic hydrogenation, gives octanoic acid. Identify A and B.
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.
Lithium has two naturally occurring isotopes: lithium−6 and lithium−7. Lithium−6 has a mass of 6.01512 relative to carbon−12 and makes up 7.47 percent of all naturally occurring lithium. Lithium−7 has a mass of 7.016 compared to carbon−12 and makes up the remaining 92.53 percent. According to this information, what is the atomic weight of lithium?
Answer:
6.941
Explanation:
Step 1:
Representation.
Let Lithium−6 be isotope A
Let Lithium−7 be isotope B
Let the abundance of isotope A (Lithium-6) be A%
Let the abundance of isotope B (Lithium-6) be B%
Step 2:
Data obtained from the question. This includes:
Mass of isotope A (Lithium−6) =
6.01512
Abundance of isotope A (Lithium−6) = A% = 7.47%
Mass of isotope B (Lithium−7) = 7.016
Abundance of isotope B (Lithium−7) = B% = 92.53%
Atomic weight of lithium =?
Step 3:
Determination of the atomic weight of lithium. This is illustrated below:
Atomic weight = [(Mass of AxA%)/100] + [(Mass of BxB%)/100]
Atomic weight = [(6.01512x7.47)/100] + [(7.016x92.53)/100]
Atomic weight = 0.449 + 6.492
Atomic weight of lithium is 6.941
Final answer:
The atomic weight of lithium is approximately 6.9423 u, which is the weighted average of the masses of its isotopes, lithium-6 and lithium-7, with respect to their natural abundances.
Explanation:
To calculate the atomic weight of lithium, we consider the relative abundances and masses of its naturally occurring isotopes, lithium-6 and lithium-7. The atomic weight is the weighted average of the atomic masses of an element's isotopes, based on their natural abundance. Using the provided information:
Lithium-6 has a mass of 6.01512 and an abundance of 7.47%.
Lithium-7 has a mass of 7.016 and an abundance of 92.53%.
The calculation is as follows:
(6.01512 × 0.0747) + (7.016 × 0.9253) = 0.4487 + 6.4936 = 6.9423
Therefore, the atomic weight of lithium is approximately 6.9423 u (atomic mass units).
Suppose the formation of iodine proceeds by the following mechanism:
step elementary reaction rate constant
1 H2 (g) + ICl (g) → HI (g) + HCl (g) k1
2 HI (g) + ICl (g) → I2 (g) + HCl (g) k2
Suppose also ki « k2. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction.
Answer:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
Overall Rate = k1[H2] [ICl]
Explanation:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.
Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence
Overall Rate = k1[H2] [ICl]
The balanced chemical equation for the overall reactions is H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g). The rate of the reaction is determined by the slowest step, in this case, the first one. So, the experimentally-observable rate law for the overall reaction would be Rate = k1[H2][ICl].
Explanation:The overall reaction is obtained by adding up the given elementary reactions. Adding step 1 and 2, we have:
H2 (g) + 2ICl (g) → 2HI (g) + HCl (g) → H2 (g) + I2 (g) + HCl (g)
After cancelling like terms the balanced chemical equation is:
H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g)
Since the first step is much slower than the second, it's the rate-determining step. The experimentally-observable rate law for the overall reaction will depend on the rate-determining step, hence, Rate = k1[H2][ICl]
Learn more about Chemical Reaction Rate here:https://brainly.com/question/34346242
#SPJ3
5. You are investigating an arson scene and you find a corpse in the rubble, but you suspect that the victim did not die as a result of the fire. Instead, you suspect that the victim was murdered earlier, and that the blaze was started to cover up the murder. How would you go about determining whether the victim died before the fire
Answer:
See the answer below.
Explanation:
Fire has three major components:
HeatSmokeGases ( in form of CO, CO2 etc)If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.
If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.
The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.
Final answer:
To determine if the victim died before the fire, forensic anthropologists analyze perimortem injuries, signs of decomposition, and toxicology for smoke inhalation indicators.
Explanation:
To determine whether a victim died before a fire, a forensic anthropologist would analyze the remains for signs of perimortem trauma. This includes examining cut marks or injuries sustained around the time of death which could indicate homicide. Disarticulation of joints and the degree of skeletal preservation can also help establish the time since death, negating the possibility that the fire caused the death if the individual was already deceased and decomposing. Moreover, the presence of smoke inhalation in the lungs can suggest if the person was alive during the fire. Toxicology reports may provide evidence of substances such as carbon monoxide or soot. The presence or absence of such substances can indicate whether death occurred before or during the fire.
Write a balanced equation for the combination reaction described, using the smallest possible integer coefficients. When carbon monoxide combines with oxygen , carbon dioxide is formed.
Answer:
2 CO(g) + O₂(g) → 2 CO₂(g)
Explanation:
Let's consider the combination reaction between carbon monoxide and oxygen to form carbon dioxide.
CO(g) + O₂(g) → CO₂(g)
We have an odd number of atoms of C on each side, so we multiply CO(g) and CO₂(g) by 2.
2 CO(g) + O₂(g) → 2 CO₂(g)
Now, the equation is balanced.
The combination reaction where carbon monoxide combines with oxygen to form carbon dioxide can be represented by the balanced chemical equation: 2CO + O₂ → 2CO₂.
Explanation:The balanced equation for the combination reaction where carbon monoxide (CO) combines with oxygen (O₂) to form carbon dioxide (CO₂) can be represented as follows:
2CO + O₂ → 2CO₂
In this reaction, two molecules of carbon monoxide react with one molecule of oxygen to produce two molecules of carbon dioxide. It's important to remember that in balancing chemical equations, one must ensure the same number of each type of atom is represented on both the reactant and product side of the equation.
Learn more about Balanced Chemical Equation here:https://brainly.com/question/28294176
#SPJ3
Describes the three-dimensional arrangement of the atoms in a molecule states that two negatively charged particles (electrons) will always repel one another with equal and opposite forces ensures bonding through shared valence electron pairs is the illustrated definition of the octet rule but not the duet rule
Answer:
When atoms other than hydrogen form covalent bonds, an octet is accomplished by sharing. The octet rule can be used to explain the number of covalent bonds an atom forms. This number normally equals the number of electrons that atom needs to have a total of eight electrons (an octet) in its outer shell
Explanation:
chemistry, the octet rule explains how atoms of different elements combine to form molecules. ... In a chemical formula, the octet rule strongly governs the number of atoms for each element in a molecule; for example, calcium fluoride is CaF2 because two fluorine atoms and one calcium satisfy the rule.
octet rule: Atoms lose, gain, or share electrons in order to have a full valence shell of eight electrons. Hydrogen is an exception because it can hold a maximum of two electrons in its valence level.
There is another rule, called the duplet rule, that states that some elements can be stable with two electrons in their shell. Hydrogen and helium are special cases that do not follow the octet rule but the duplet rule. ... They are stable in a duplet state instead of an octet state.
Why would you use a solution, such as a cabbage pH indicator, to measure the pH of household items?
a) to see if food went bad
b) to test the safety of water
c) to make sure conditions are safe
d) to use up old cabbage
Answer:
-to see if food went bad
-to test the safety of water
-to make sure conditions are safe
Explanation:
Answer:
A, B, C
Explanation:
Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met, phe, ser, and tyr. carboxypeptidase a releases ser. edman's reagent releases leu. treatment with cyanogen bromide forms two peptides with the following amino acid compositions: 1. arg, phe, ser 2. arg, leu, lys, met, tyr trypsin-catalyzed hydrolysis forms the following two amino acids and two peptides: 1. arg 2. ser 3. arg, met, phe 4. leu, lys, tyr
A Sentence that I Used:
The position of Met seems incorrect. The amino acid sequences of all proteins begin with Met because it is the amino acid that is attached to the anticodon for the AUG start codon. It looks like the student might have worked in a backward direction for transcription.
A precipitate forms when mixing solutions of silver nitrate (AgNO,) and sodium chloride (NaCI). Complete the net ionic equation for this reaction by filling in the blanks. Do not include charges on any ions and do not include phase symbols. Provide your answer below (aq)+(aq)6)
The reaction between solutions of silver nitrate and sodium chloride results in a precipitate due to a double-replacement reaction. The solid precipitate formed is silver chloride (AgCl). The net ionic equation becomes: Ag+ (aq) + Cl- (aq) → AgCl (s).
Explanation:
When you mix solutions of silver nitrate (AgNO3) and sodium chloride (NaCl), a precipitate forms. This is because of a double-replacement reaction that occurs between the two compounds. You start with AgNO3 (silver nitrate) and NaCl (sodium chloride), and end up with NaNO3 (sodium nitrate) and AgCl (silver chloride). The silver chloride precipitates, or becomes solid, leaving the sodium nitrate in solution. This can be shown by the net ionic equation:Ag+ (aq) + Cl- (aq) → AgCl (s). This equation represents the formation of the silver chloride precipitate. The ions Na+ and NO3- do not take part in the reaction and so are not included in the net ionic equation.
Learn more about Net Ionic Equation here:https://brainly.com/question/35304253
#SPJ12
What amount of solid NaOH must be added to 1.0 L of a 0.14 M H2CO3 solution to produce a solution with [H+]= 3.5×10−11 M ? There is no significant volume change as the result of the addition of the solid.
To produce a solution with [H+]= 3.5×10−11 M, 0.14 mol of solid NaOH must be added to 1.0 L of a 0.14 M H2CO3 solution.
Explanation:To calculate the amount of solid NaOH needed, we need to use the balanced chemical equation for the reaction between NaOH and H2CO3. The equation is:
NaOH + H2CO3 -> Na2CO3 + H2O
From the equation, we can see that the ratio between NaOH and H2CO3 is 1:1. Therefore, the amount of solid NaOH needed is equal to the molarity of H2CO3 multiplied by the volume, which is:
Amount of solid NaOH = 0.14 M * 1.0 L = 0.14 mol NaOH
Learn more about Molarity here:https://brainly.com/question/8732513
#SPJ11
Electrospray ionization and atmospheric pressure chemical ionization are two methods that are used to introduce the eluate from the liquid chromatography column into a mass spectrometer. Which ionization method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer?
Answer:
Electrospray ionization
Explanation:
Electrospray ionization Is a soft ionization technique used in producing ions from macromolecules, it is employed especially in spectrometry where high voltage is used to create an aerosol from a liquid.It can be employed in Knowing molecular weights of molecules and biological macromolecules such as Peptides and proteins. Therefore, electrospray ionization is the method typically requires analyte ions to be in solution prior to reaching the interface between the column in liquid chromatography and the mass spectrometer