Explanation:
It is known that for high concentration of [tex]M^{2+}[/tex], reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.
Now, [tex]E^{o}_{cell}[/tex] = 0 and the general reaction equation is as follows.
[tex]M^{2+} + M \rightarrow M + M^{2+}[/tex]
3.00 M n = 2 30 mM
E = [tex]0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}[/tex]
= [tex]-\frac{0.0591}{2} log (5 \times 10^{-2})[/tex]
= 0.038 V
Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.
Ethylene () is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane () from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas. Suppose an engineer studying ethane cracking fills a reaction tank with of ethane gas and raises the temperature to . He believes at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. Round your answer to significant digits. Note for advanced students: the engineer may be mistaken about the correct value of , and the mass percent of ethylene you calculate may not be what he actually observes.
The question is incomplete, here is the complete question:
Ethylene is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10¹⁰ of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas.
Suppose an engineer studying ethane cracking fills a 30.0 L reaction tank with 38.0 atm of ethane gas and raises the temperature to 400°C. He believes Kp = 0.4 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture.
Answer: The mass percent of ethylene gas is 9.20 %
Explanation:
We are given:
Initial partial pressure of ethane gas = 38.0 atm
The chemical equation for the dehydrogenation of ethane follows:
[tex]C_2H_6\rightleftharpoons C_2H_4+H_2[/tex]
Initial: 38
At eqllm: 38-x x x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]
We are given:
[tex]K_p=0.40[/tex]
Putting values in above equation, we get:
[tex]0.40=\frac{x\times x}{38-x}\\\\x=-4.10,3.70[/tex]
Neglecting the negative value of 'x' because partial pressure cannot be negative
So, equilibrium partial pressure of ethane = 38 - x = 38 - 3.70 = 34.30 atm
Equilibrium partial pressure of ethene = x = 3.70 atm
To calculate the number of moles, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex] ..........(1)
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(2)
For ethane:We are given:
[tex]P=34.3atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]
Putting values in equation 1, we get:
[tex]34.3atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{34.3\times 30.0}{0.0821\times 673}=18.62mol[/tex]
Molar mass of ethane gas = 30 g/mol
Moles of ethane gas = 18.62 mol
Putting values in equation 2, we get:
[tex]18.62mol=\frac{\text{Mass of ethane}}{30g/mol}\\\\\text{Mass of ethane gas}=(18.62mol\times 30g/mol)=558.6g[/tex]
For ethylene:We are given:
[tex]P=3.70atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]
Putting values in equation 1, we get:
[tex]3.70atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{3.70\times 30.0}{0.0821\times 673}=2.01mol[/tex]
Molar mass of ethylene gas = 28 g/mol
Moles of ethylene gas = 2.01 mol
Putting values in equation 2, we get:
[tex]2.01mol=\frac{\text{Mass of ethylene}}{28g/mol}\\\\\text{Mass of ethylene gas}=(2.01mol\times 28g/mol)=56.28g[/tex]
To calculate the mass percentage of substance in mixture, we use the equation:[tex]\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100[/tex]
Mass of ethylene = 56.28 g
Mass of mixture = [558.6 + 56.28] g = 641.88 g
Putting values in above equation, we get:
[tex]\text{Mass percent of ethylene}=\frac{56.28g}{641.88g}\times 100=9.20\%[/tex]
Hence, the mass percent of ethylene gas is 9.20 %
Consider the following four titrations: i. 100.0 mL of 0.10 M HCl titrated with 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated with 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated with 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated with 0.10 M NaOH Rank the titrations in order of increasing volume of titrant added to reach the equivalence point.
Explanation:
As we know that HCl is a stronger acid and NaOH is a stronger base.
And, HF and phenol are weaker acids having [tex]K_{a}[/tex] values of [tex]6.6 \times 10^{-4}[/tex] and [tex]1.3 \times 10^{-10}[/tex] respectively.
In the same way, methyl amine and pyridine are weaker bases with [tex]K_{b}[/tex] values of [tex]4.4 \times 10^{-4}[/tex] and [tex]1.7 \times 10^{-9}[/tex] respectively.
(i) Volume will be calculated as follows.
[tex]M_{a}V_{a} = M_{b}V_{b }[/tex]
[tex]100 \times 0.1 = 0.1 \times V[/tex]
Therefore, volume of NaOH is 100 ml.
(ii) Similarly, volume of HCl is 100 ml.
(iii) For Methyl amine,
[tex][OH]^{-} = \sqrt{4.4 \times 10^{-4} \times 0.1}[/tex]
= [tex]6.6 \times 10^{-3}[/tex]
And as, [tex]M_{a}V_{a} = M_{b}V_{b}[/tex]
[tex]0.1 \times V = 6.6 \times 10^{-3} \times 100[/tex]
Hence, the volume of HCl is 6.6 ml.
(iv) For HF,
[tex][H]^{+} = \sqrt{6.6 \times 10^{-4} \times 0.1}[/tex]
= [tex]8.12 \times 10^{-3}[/tex]
As, [tex]M_{a}V_{a} = M_{b}V_{b}[/tex]
[tex]8.12 \times 10^{-3} \times 100 = 0.1 \times V[/tex]
Hence, the volume of NaOH added is 8.12 ml.
Therefore, we can conclude that the increasing order of volums of given titrant is (i) = (ii) > (iv) > (iii).
All titrations (i to iv) involve acid-base reactions with equal molarity and volume of reactants; therefore, the volume of titrant at the equivalence point is the same for each scenario, which is 100.0 mL.
Explanation:To rank the titrations in order of increasing volume of titrant added to reach the equivalence point, we must consider the nature of the reactants in each titration. Strong acids and bases will react in a 1:1 ratio, meaning an equal volume of titrant is needed to reach the equivalence point if their concentrations are the same. Thus, titrations i and ii, which involve the strong acid HCl and the strong base NaOH, will have the same volume of titrant at the equivalence point.
Titeration iii, which involves the weak base CH₃NH₂ and the strong acid HCl, will have a different volume compared to a strong acid/strong base titration due to the possibility of incomplete dissociation of the weak base. However, since the concentrations are equal, 100.0 mL of titrant is still expected to be needed to reach the equivalence point.
Titeration iv, which includes the weak acid HF and strong base NaOH, also involves a 1:1 stoichiometry at the equivalence point, but weak acids can exhibit buffering effects which may slightly alter the volume needed to reach the equivalence point. Nevertheless, with equal concentrations, the same volume of titrant is expected to be used as in the previous cases.
Therefore, because all titrations have equal molarities and volumes of both the titrants and analytes, we'd expect the volume of titrant needed to reach the equivalence point to be the same for each scenario, which is 100.0 mL.
In order to safely run a sample in a centrifuge, it is important to Select one: a. balance the sample with a counterweight of the same mass directly across from it. b. make sure the sample is free from impurities. c. ensure that the sample fills the centrifuge tube completely. d. do all of the given tasks.
Answer: Option (b) is the correct answer.
Explanation:
A laboratory device which is used to separate fluids, gases or liquid on the basis of their density is known as a centrifuge. When a vessel containing the sample is spins at a high speed then separation takes place as the centrifugal force pushes the heavier material out of the vessel during this process.
So, if we want the sample to safely run through the centrifuge then it is important that it should be free from impurities.
Thus, we can conclude that in order to safely run a sample in a centrifuge, it is important to make sure the sample is free from impurities.
When the safety run sample should be in a centrifuge so the sample should be free from impurities.
Laboratory device:It is used to distinguish gases or liquids that depend upon the density is called a centrifuge. At the time When a vessel comprised of the sample should be spinning at a high speed so the separation took place since the centrifugal force pushes the heavier material that should be out of the vessel at the time of the process.
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A chemist prepares a solution of pottasium bromide KBr by measuring out 224.g of pottasium bromide into a 300.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mo/L of the chemist's pottasium bromide solution. Round your answer to 3 significant digits.
Explanation:
Below is an attachment containing the solution.
Answer:
The concentration of the KBr solution is 6.27 mol/L
Explanation:
Step 1: Data given
Mass of KBr = 224 grams
Molar mass KBr = 119.0 g/mol
Volume = 300 mL = 0.300 L
Step 2: Calculate moles KBr
Moles KBr = mass KBr / molar mass KBr
Moles KBr = 224 grams / 119.0 g/mol
Moles KBr = 1.88 moles
Step 3: Calculate concentration KBr solution
Concentration solution = moles KBr / volume
Concentration solution = 1.88 moles / 0.300 L
Concentration solution = 6.27 mol / L
The concentration of the KBr solution is 6.27 mol/L
1. The average annual increase of CO2 between 1958 and 2014 was . 2. The average annual increase of CO2 between 1970 and 1980 was . 3. The average annual increase of CO2 between 2004 and 2014 was . 4. CO2 concentration and rate over time. 5. The most recent CO2 concentration measured is .
The CO2 concentration has been increasing steadily for several decades, with an average annual increase about 1.4 ppm between 1958 and 2014, somewhat lower between 1970 and 1980 and higher between 2004 and 2014. The most recent CO2 concentration measured I found is 414.0 ppm up to July 2021.
Explanation:It appears that the student's question is missing specific data, however, I can provide information on the average increase in
CO2 concentration
over certain periods. The concentration of CO2 in the atmosphere has been steadily increasing since the beginning of the industrial revolution, with more prominent increases in recent decades due to increased industrial activity and deforestation. For example, between 1958 and 2014, the CO2 concentration in the atmosphere reportedly increased by about 1.4 parts per million (ppm) per year on average. The rate was somewhat lower between 1970 and 1980 (about 1.3 ppm/year), but higher between 2004 and 2014 (about 2 ppm/year). The most recent CO2 concentration recorded I could find is 414.0 ppm up to July 2021. It's important to note that
increasing CO2 concentration
is a significant contributor to global warming and climate change.
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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 given that the Ka of C6H5COOH is 6.5×10-5 and the molar mass of NaC6H5COO is 144.1032 g/mol?
Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
Draw the two constitutionally isomeric structures formed when iodobenzene and propene are subjected to the conditions of the Heck reaction. If stereoisomers are possible for a particular constitutional isomer, draw the more stable stereoisomer.
Answer:
Check the explanation
Explanation:
The diagram to the question is in the attached image, and always remember that Constitutional isomers are compounds that have very similar molecular formula and diverse connectivity. To conclude whether two molecules are constitutional isomers, just calculate the figure or amount of every atom in both molecules and see how the atoms are assembled.
What is the enthalpy change (in kJ) of a chemical reaction that raises the temperature of 250.0 ml of solution having a density of 1.25 g/ml by 7.80°C? (The specific heat of the solution is 3.74 joules/gramK.) A. -7.43 kJ
Answer:
328.4KJ
Explanation:
Before we move on to calculate enthalpy change, we calculate the amount of heat Q
Q= mcΔT
m = density * volume = 250 * 1.25 = 312.5g
c = 3.74J/g.k
ΔT = 7.80 + 273.15K = 280.95K
Q= 312.5 * 3.74 * 280.95 = 328,360.312 J= 328.4KJ(1000J = 1KJ, so divide by 1000)
The enthalpy change in the reaction is same as amount of heat transferred = 328.4KJ
A piece of copper has a mass of 800 g. What is the volume of the sample, in units of liters? In the boxes above, enter the correct setup that would be used to solve this problem.
Answer:
0.089L
Explanation:
Mass of copper= 800g
Density of copper= 8.96g/ml or 8960g/L
Density = mass/volume
Volume = mass/density = 800/8960= 0.089L
The conversion of density from g/ml to g/L units was necessary because the volume was required in liters according to the statement in the question
Consider the following mechanism: (1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast] (2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow] (3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast] (a) What is the overall equation? Select the single best answer. ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) → IO−(aq) + Cl−(aq) (b) Identify the intermediate(s), if any. Select the single best answer. No intermediates Cl−, OH−, I−, ClO−, IO− HClO, OH−, HIO HClO, OH−, HIO, H2O (c) What are the molecularity and the rate law for each step? Select the single best answers. (1): bimolecular unimolecular termolecular rate = k1([HClO][OH−])/([HClO][OH−]) k1[HClO][OH−] k1[ClO−][H2O] (2): bimolecular unimolecular termolecular rate = k2[HIO][Cl−] k2([HIO][Cl−])/([I−][HClO]) k2[I−][HClO] (3): bimolecular unimolecular termolecular rate = k3[OH−][HIO] k3([H2O][IO−])/([OH−][HIO]) k3[H2O][IO−] (d) Is the mechanism consistent with the actual rate law: rate = k[ClO−][I−]? no yes
Answer:
1. The overall equation is ClO-(aq)+I-(aq) → Cl-(aq)+IO-(aq)
2. The intermediates include: HClO(aq), OH-(aq) and HIO(aq)
3. The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]
4. No,
The rate depends on [OH-], so it's not consistent with the actual rate law
Explanation:
1
Given
(1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]
(2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]
(3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]
Add up the three equations
ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]
I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]
OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]
Remove all common terms {H2O(l) + I-(aq) +HClO(aq) +OH-(aq) +HIO(aq) => HClO(aq) + OH-(aq) + HIO(aq)
+H2O(l)}
We're left with
ClO-(aq) + I-(aq) => Cl-(aq) + IO-(aq)
2.
There are intermediates generated but they are not visible in the overall equation.
The intermediates include: HClO(aq), OH-(aq) and HIO(aq)
3.
The three steps are bimolecular.
The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]
4. Let K represents equilibrium constant
At step 1,
K1 = [HClO][OH-]/[ClO-]
Simplify;
K1 [ClO-]= [HClO][OH-]
K1[ClO-]/[OH-] = [HClO]
Determine the rate at step 2
= k2[I-][HClO]
= K1k2[I-][ClO-]/[OH-]
= k[ClO-][I-]/[OH-]
The answer is no
Final answer:
The overall equation derived from the given mechanism is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq), with HClO, OH−, and HIO as intermediates. Molecularity for all steps is bimolecular with specific rate laws for each step, and the mechanism is consistent with the experimental rate law rate = k[ClO−][I−].
Explanation:
The student's question pertains to deriving the overall equation, identifying intermediates, determining molecularity and rate laws for each step, and verifying the consistency of a proposed mechanism with the experimental rate law in a chemical reaction series involving species such as ClO−(aq), H2O(l), I−(aq), and others.
Answers to the Student's Question
The overall equation is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq).
The intermediates in the reaction mechanism are HClO, OH−, HIO.
For the first step, the molecularity is bimolecular and the rate law is k1[ClO−][H2O]. For the second step, it's bimolecular with a rate law of k2[I−][HClO]. The third step is also bimolecular with a rate law of k3[OH−][HIO].
The mechanism is consistent with the actual rate law, which is rate = k[ClO−][I−], because the rate-determining step involves these reactants.
Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant. (a) 11.00 mL: pH = (b) 20.60 mL: pH = (c) 25.00 mL:
The pH changes significantly through the different stages of titration of triethylamine with HCl. After adding 11.00 mL of HCl, the pH is approximately 11.30, demonstrating a basic solution. At 20.60 mL and 25.00 mL of HCl added, the pH drops to approximately 2.83 and 1.95, respectively, indicating an acidic solution due to the excess HCl.
To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, [((CH₃CH₂)₃N)] , with 0.1000 M HCl solution, we need to follow the steps below at different volumes of HCl addition:
(a) After adding 11.00 mL of HCl
Calculate moles of triethylamine:
moles of [((CH₃CH₂)₃N)] = 0.1000 M x 0.0200 L = 0.00200 mol
Calculate moles of HCl added:
moles of HCl = 0.1000 M * 0.0110 L = 0.00110 mol
Moles of triethylamine remaining:
0.00200 mol - 0.00110 mol = 0.00090 mol
Volume of solution after adding HCl:
20.00 mL + 11.00 mL = 31.00 mL or 0.0310 L
Concentration of triethylamine:
[((CH₃CH₂)₃N)] = 0.00090 mol / 0.0310 L ≈ 0.0290 M
Using Kᵇ, find [OH⁻]:
Kᵇ = 5.2 x 10⁻⁴, set up ICE table for equilibrium calculation, estimate [OH⁻].
Find pOH and then pH:
pOH = -log[OH⁻]; pH = 14 - pOH
pH ≈ 11.30
(b) After adding 20.60 mL of HCl
Moles of HCl added:
0.1000 M x 0.0206 L = 0.00206 mol
Moles of triethylamine remaining:
0.00200 mol - 0.00206 mol = -0.00006 mol. This indicates an excess of HCl.
Moles of excess HCl:
0.00006 mol
Volume of solution:
20.00 mL + 20.60 mL = 40.60 mL or 0.0406 L
Concentration of H⁺:
[H⁺] = 0.00006 mol / 0.0406 L ≈ 0.00148 M
Find pH:
pH = -log[H⁺]
pH ≈ 2.83
(c) After adding 25.00 mL of HCl
Moles of HCl added:The pH values during the titration are (a) 10.65 after adding 11.00 mL of HCl, (b) 3.97 at equivalence point (20.60 mL), and (c) 1.95 after adding 25.00 mL of HCl.
To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH₃CH₂)₃N (Kb = 5.2 × 10⁻⁴), with 0.1000 M HCl solution after different volumes of titrant have been added, follow these steps:
Initial pH Calculation (before titration)
1. Calculate the pOH from Kb and the initial concentration of triethylamine:
Kb = 5.2 × 10⁻⁴
[B] = 0.1000 M
Using the formula:
[tex]& K_b = \frac{[\text{OH}^-][\text{BH}^+]}{[B]} \\[/tex]
We assume that [OH⁻] = [BH⁺]. Thus, [tex]K_b = \frac{[\text{OH}^-]^2}{[B]} \\[/tex]
[tex]& [\text{OH}^-] = \sqrt{K_b \cdot [B]} = \sqrt{5.2 \times 10^{-4} \cdot 0.1000} = 0.0072 \, \text{M} \\\\[/tex]
[tex]& \text{pOH} = -\log[\text{OH}^-] = -\log(0.0072) \approx 2.14 \\\\ & \text{pH} = 14 - \text{pOH} = 14 - 2.14 = 11.86 \\[/tex]
After Adding 11.00 mL of HCl
2. Calculate the moles of HCl added:
n(HCl) = M * V = 0.1000 M * 0.01100 L = 0.001100 mol
Initial moles of (CH₃CH₂)₃N = 0.1000 M * 0.02000 L = 0.002000 mol
Remaining (CH₃CH₂)₃N = 0.002000 mol - 0.001100 mol = 0.000900 mol
Moles of (CH₃CH₂)₃NH⁺ formed = 0.001100 mol
Total volume = 20 mL + 11 mL = 31 mL = 0.031 L
Compute the concentrations:
[B] = 0.000900 mol / 0.031 L ≈ 0.029 M
[HB+] = 0.001100 mol / 0.031 L ≈ 0.035 M
Using the Henderson-Hasselbalch equation:
[tex]pH = pKa + \log\left(\frac{[B]}{[\text{HB}^+]}\right)\\\\pKa = 14 - \text{pKb} = 14 - 3.28 = 10.72\\\\pH = 10.72 + \log\left(\frac{0.029}{0.035}\right) = 10.72 + \log(0.829) \approx 10.65[/tex]
At Equivalence Point (20.60 mL)
3. Calculate moles at equivalence point:
Moles of (CH₃CH₂)₃N = 0.002000 mol
Equivalence moles of HCl = 0.002060 mol
All the base has been neutralized:
(CH₃CH₂)₃NH⁺ in solution = 0.002060 / 0.04060 = 0.0507 M
Calculate pH:
[H+] from the equilibrium of (CH₃CH₂)₃NH⁺ :
[tex]K_a \text{ for } (CH_3CH_2)_3NH^+ = \frac{1}{K_b} = \frac{1}{5.2 \times 10^{-4}} = 1.923 \times 10^3[/tex]
[tex]& [\text{H}^+] = \sqrt{1.923 \times 10^3 \times 0.0507 \, \text{M}} = 0.09825 \, \text{M} \\\\& \text{pH} = -\log(0.09825) \approx 3.97 \\[/tex]
After Adding 25.00 mL of HCl
4. Calculate moles of excess HCl:
HCl added (total) = 0.002500 moles
Excess HCl = 0.002500 - 0.002000 = 0.000500 mol
Total volume = 20 mL + 25 mL = 45 mL = 0.045 L
[tex]& [\text{H}^+] = \frac{0.000500 \, \text{mol}}{0.045 \, \text{L}} \approx 0.0111 \, \text{M} \\[/tex]
[tex]& \text{pH} = -\log(0.0111) \approx 1.95 \\[/tex]
In summary, the pH values during the titration are 10.65 after adding 11.00 mL of HCl, 3.97 at equivalence point (20.60 mL), and 1.95 after adding 25.00 mL of HCl.
High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 6% of all the original hydrogen atoms.
Explanation:
Let us assume that 50 carbon atoms are available with possible 100 site bondings. It is given here that 94% are occupied by hydrogen (94 out of 100) and 6% (6 out of 100) are occupied by chlorine atom.
Therefore, moles of carbon = 50
moles of hydrogen = 94
moles of chlorine = 6
Therefore,
Mass of 50 carbon atoms = [tex]50 \times 12.01 g/mol[/tex] = 600.5 g/mol
Mass of 94 hydrogen atoms = [tex]94 \times 1.008 g/mol[/tex] = 94.752 g/mol
Mass of 6 chlorine atoms = [tex]6 \times 35.45 g/mol[/tex] = 212.7 g/mol
Therefore, concentration of chlorine is as follows.
[tex]C_{cl} = \frac{m_{cl}}{m_{c} + m_{H} + m_{cl}}[/tex]
= [tex]\frac{212.7}{907.952} \times 100[/tex]
= 23.42%
thus, we can conclude that the concentration of Cl is 23.42%.
The calculation of Chlorine concentration in high-density polyethylene after 6% Hydrogen replacement involves determining the increased mass from Chlorine substitution, considering the significant difference in atomic masses of Hydrogen and Chlorine. Chlorine's contribution to the weight of the new compound would be considerable.
Explanation:High-density polyethylene has a chemical formula of (C2H4)n, where 'n' denotes the number of repeating units in the polymer chain. Each molecule of ethylene contributes two hydrogen atoms. Therefore, if we assume 6% of all hydrogen atoms are replaced by chlorine atoms, we have to add the weight of chlorine into the composition.
Under normal conditions, Chlorine (Cl) has an atomic weight of approximately 35.45 g/mol, and Hydrogen (H) has an atomic weight of approximately 1.008 g/mol. Therefore, the introduction of Chlorine substitutes a weight of 35.45 g for every 1.008 g of Hydrogen.
To calculate the weight percent of Chlorine in the composition, we determine the total weight contributed by chlorine and divide it by the total weight of the new compound, then multiply by 100%. As a result, the weight percentage of Chlorine when 6% of hydrogen atoms are replaced, following the calculations and the Principle of Atomic Substitution, is quite significant due to the considerable difference in atomic weights of Chlorine and Hydrogen.
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What is the total pressure (in atm) inside of a vessel containing N2 exerting a partial pressure of 0.256 atm, He exerting a partial pressure of 203 mmHg, and H2 exerting a partial pressure of 39.0 kPa?
Answer: Total pressure inside of a vessel is 0.908 atm
Explanation:
According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.
[tex]p_{total}=p_1+p_2+p_3[/tex]
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 0.256 atm
[tex]p_{He}[/tex] = partial pressure of helium = 203 mm Hg = 0.267 atm (760mmHg=1atm)
[tex]p_{H_2}[/tex] = partial pressure of hydrogen =39.0 kPa = 0.385 atm (1kPa=0.00987 atm)
Thus [tex]p_{total}=p_{H_2}+p_{He}+p_{H_2}[/tex]
[tex]p_{total}[/tex] =0.256atm+0.267atm+0.385atm =0.908atm
Thus total pressure (in atm) inside of a vessel is 0.908
Answer:
The total pressure inside the vessel is 0.908 atm
Explanation:
Step 1: Data given
Partial pressure N2 = 0.256 atm
Partial pressure He = 203 mmHg = 0.267105 atm
Partial pressure H2 = 39.0 kPa = 0.3849 atm
Step 2: Calculate the total pressure
Total pressure = p(N2) + p(He) + p(H2)
Total pressure = 0.256 atm + 0.267105 atm + 0.3849 atm
Total pressure = 0.908 atm
The total pressure inside the vessel is 0.908 atm
A breeder nuclear reactor is a reactor in which nonfissile U-238 is converted into fissile Pu-239. The process involves bombardment of U-238 by neutrons to form U-239 which then undergoes two sequential beta decays. Write nuclear equations to represent this process.
Answer: The nuclear equations for the given process is written below.
Explanation:
The chemical equation for the bombardment of neutron to U-238 isotope follows:
[tex]_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}[/tex]
Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the first beta decay process of [tex]_{92}^{239}\textrm{U}[/tex] follows:
[tex]_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta[/tex]
The chemical equation for the second beta decay process of [tex]_{93}^{239}\textrm{Np}[/tex] follows:
[tex]_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta[/tex]
Hence, the nuclear equations for the given process is written above.
Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl4(g)] = −96.0 kJ/mol ΔH o f [CCl4(l)] = −139 kJ/mol ΔH o f [HCl(g)] = −92.31 kJ/mol ΔH o f [HCl(aq)] = −167.46 kJ/mol ΔH o f [Cl(g)] = 121.0 kJ/mol
To calculate the standard enthalpy change for the given chemical reaction, balance the equation, then apply standard enthalpies of formation for the reactants and products, and use Hess's Law. The calculated standard enthalpy change for this reaction is -433.37 kJ/mol.
To calculate the standard enthalpy change ([tex]H_orxn[/tex]) for the reaction [tex]CH_4(g) + Cl_2(g) \rightarrow CCl_4(l) + HCl(g)[/tex], we need to first balance the reaction:
[tex]CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)[/tex]
Then we use the standard enthalpies of formation ( Hf) for each substance to calculate [tex]H_orxn[/tex] using Hess's Law:
[tex]H_o = [ H_f[CCl_4(l)] + 4 H_f[HCl(g)]] - [ H_f[CH_4(g)] + 4 H_f[Cl_2(g)]][/tex]
[tex]H_o = [-139 kJ/mol + 4(-92.31 kJ/mol)] - [-74.87 kJ/mol + 4(0 kJ/mol)][/tex]
[tex]H_o = -139 kJ/mol - 369.24 kJ/mol - (-74.87 kJ/mol)[/tex]
[tex]H_o = -433.37 kJ/mol[/tex]
ΔH°rxn for CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), is -433.37 kJ
To calculate the enthalpy change (ΔH°rxn) for the reaction CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), we first need to balance the equation and use the standard enthalpies of formation (ΔH°f) provided for each compound.
Balance the chemical equation:
The balanced equation is: CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g)
Write the enthalpy of formation values:
[tex]\Delta H^\circ_f[\text{CH}_4(g)] &= -74.87 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{CCl}_4(\ell)] &= -139 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{HCl(g)}] &= -92.31 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{Cl}_2(g)] &= 0 \, \text{kJ/mol} \quad (\text{by convention})[/tex]
Calculate the total ΔH°f for the products:
[tex]\text{For CCl}_4(\ell) \text{ and HCl(g):}\\\\ & {1 \, \text{mol of CCl}_4(\ell) \times (-139 \, \text{kJ/mol})} + {4 \, \text{mol of HCl(g)} \times (-92.31 \, \text{kJ/mol})} \\& = -139 \, \text{kJ} + (-369.24 \, \text{kJ}) \\& = -508.24 \, \text{kJ} \\[/tex]
Calculate the total ΔH°f for the reactants:
[tex]\text{For CH}_4(g) \text{ and Cl}_2(g): \\\\& {1 \, \text{mol of CH}_4(g) \times (-74.87 \, \text{kJ/mol})} + {4 \, \text{mol of Cl}_2(g) \times (0 \, \text{kJ/mol})} \\& = -74.87 \, \text{kJ} \\[/tex]
Determine ΔH°rxn:
[tex]\Delta H^\circ_\text{rxn}: & = \Sigma \Delta H^\circ_f(\text{products}) - \Sigma \Delta H^\circ_f(\text{reactants}) \\\\& = -508.24 \, \text{kJ} - (-74.87 \, \text{kJ}) \\& = -433.37 \, \text{kJ}[/tex]
Therefore, the enthalpy change for the reaction ΔH°rxn is -433.37 kJ.
Did you notice a difference in the number of times your substrate underwent addition in the two Friedel-Crafts reactions? Why was there a difference or why was there no difference?
Answer:
a. Yes, there was a difference in the number of times your substrate underwent addition in the two Friedel-Crafts reactions.
b. There was a difference- due to the fact that the first added alkyl group increaseD the electron density on benzene ring.
Explanation:
Yes, there was a difference in the number of times the substrate underwent addition in the two Friedel-Crafts reactions .
There was a difference - There was an increase in the rate of second Friedel-Craft reaction due to the fact that the first added alkyl group increaseD the electron density on benzene ring.
Therefore the rate of the addition reaction increases.
Friedel-Crafts reactions exhibit differences in substrate addition frequency due to the presence of electron-donating alkyl groups, which increase reactivity through polyalkylation, and activating groups that stabilize intermediates and affect reactivity.
The differences in the number of times your substrate underwent addition in the two Friedel-Crafts reactions can be attributed to the nature of the substituents added during the process. In the Friedel-Crafts alkylation, alkyl groups are introduced onto an aromatic ring which are electron-donating substituents. These alkyl groups render the product more reactive than the starting material, leading to the possibility of polyalkylation, where multiple alkyl groups can be added if additional reaction takes place. This is due to the alkyl groups' ability to activate the ring towards further electrophilic attack. Polyalkylation often results in a mixture of products and for synthetic purposes, this could be considered a disadvantage as it complicates purification and affects yield.
On the other hand, the presence of an activating group, such as a methoxy substituent, in Friedel-Crafts alkylation reactions can stabilize the ring carbocation intermediate and increase the rate of reaction, leading to a difference in reactivity compared to reactions without such group. Thus, the control of reactivity and selectivity in Friedel-Crafts reactions is critical and differences in the substrate or reaction conditions can have a significant impact on the outcome of the reaction.
Spray drying_______________.a. converts solid foods to semi-solid foods b. atomizes liquids into small solid particles c. increases moisture content d. is similar to fluidized bed drying
Answer:
b. atomizes liquids into small solid particles
Explanation:
Spray drying -
It refers to the process of getting a dry powder from the liquid or the slurry ,with the help of hot gas , is referred to as spray drying .
The method is used in the field of pharmacy .
It is a type of atomizer or spray nozzle which helps to disperse the liquid into the controlled drop size spray .
Small solid particles are generated by this method .
Hence , from the given scenario of the question ,
The correct answer is b. atomizes liquids into small solid particles .
Calculate the amount of heat needed to melt 91.5g of solid benzene ( C6H6 ) and bring it to a temperature of 60.6°C . Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer: The amount of heat required for melting of benzene is 20.38 kJ
Explanation:
The processes involved in the given problem are:
[tex]1.)C_6H_6(s)(5.5^oC)\rightleftharpoons C_6H_6(l)(5.5^oC)\\2.)C_6H_6(l)(5.5^oC)\rightleftharpoons C_6H_6(l)(60.6^oC)[/tex]
For process 1:To calculate the amount of heat required to melt the benzene at its melting point, we use the equation:
[tex]q_1=m\times \Delta H_{fusion}[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
m = mass of benzene = 91.5 g
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 127.40 J/g
Putting all the values in above equation, we get:
[tex]q_1=91.5g\times 127.40J/g=11657.1J[/tex]
For process 2:To calculate the amount of heat absorbed at different temperature, we use the equation:
[tex]q_2=m\times C_{p}\times (T_{2}-T_{1})[/tex]
where,
[tex]C_{p}[/tex] = specific heat capacity of benzene = 1.73 J/g°C
m = mass of benzene = 91.5 g
[tex]T_2[/tex] = final temperature = 60.6°C
[tex]T_1[/tex] = initial temperature = 5.5°C
Putting values in above equation, we get:
[tex]q_2=91.5\times 1.73J/g^oC\times (60.6-(5.5))^oC\\\\q_2=8722.05J[/tex]
Total heat absorbed = [tex]q_1+q_2[/tex]
Total heat absorbed = [tex][11657.1+8722.05]J=20379.2=20.38kJ[/tex]
Hence, the amount of heat required for melting of benzene is 20.38 kJ
Final answer:
The amount of heat produced by the combustion of the benzene sample is 6.562 kJ.
Explanation:
To calculate the amount of heat produced by the combustion of the benzene sample, we can use the heat capacity of the bomb calorimeter and the change in temperature. The heat produced can be calculated using the formula: q = C × ΔT, where q is the heat produced, C is the heat capacity of the bomb calorimeter, and ΔT is the change in temperature. In this case, the heat capacity of the bomb calorimeter is 784 J/°C and the change in temperature is 8.39 °C.
First, we calculate the heat produced by the bomb calorimeter using the formula: q = C × ΔT. q = 784 J/°C × 8.39 °C = 6561.76 J. Next, we convert this to kilojoules by dividing by 1000: 6561.76 J ÷ 1000 = 6.562 kJ.
Therefore, the amount of heat produced by the combustion of the benzene sample is 6.562 kJ.
Calculate and report the precise concentration of undiluted stock standard solution #1 for AR in micromoles per liter from ppm by mass. Assume that the density of water is 1.00g/ml. This is your most concentrated undiluted standard solution for which you measured the absorbance.
The question is incomplete, complete question is ;
Allura Red (AR) has a concentration of 21.22 ppm. What is this is micro moles per liter? Report the precise concentration of the undiluted stock solution #1 of AR in micromoles per liter. This is your most concentrated (undiluted) standard solution for which you measured the absorbance. Use 3 significant figures. Molarity (micro mol/L) =
Answer:
The molarity of the solution of allura red is 42.75 micro moles per Liter.
Explanation:
The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.
To calculate the ppm of oxygen in sea water, we use the equation:
[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]
Both the masses are in grams.
We are given:
The ppm concentration of allura red = 21.22 ppm
This means that 21.22 mg of allura red was present 1 kg of solution.
Mass of Allura red = 21.22 mg = [tex]21.22\times 0.001 g[/tex]
1 mg = 0.001 g
Mass of solution = 1 kg = 1000 g
Density of the solution = Density of water = d = 1.00 g/mL
( since solution has very small amount of solute)
Volume of the solution :
[tex]=\frac{1000 g}{1.00 g/mL}=1000 mL[/tex]
1000 mL = 1 L
Volume of the solution, V = 1 L
Moles of Allura red = [tex]\frac{21.22\times 0.001 g}{496.42 g/mol}=4.275\times 10^{-5} mol=4.275\times 10^{-5}\times 10^{6} \mu mole[/tex]
Molarity of the solution ;
[tex]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}[/tex]
[tex]M=\frac{4.275\times 10^{-5}\times 10^6 \mu mol}{1 L}=42.75 \mu mol/L[/tex]
The molarity of the solution of allura red is 42.75 micro moles per Liter.
The Fischer esterification mechanism is examined in the following two questions in the assignment. Part 1 involves MeOH addition to form the key tetrahedral intermediate. Part 2 will involve loss of H2O to form the ester. Part 1 of 2:
The Fischer esterification mechanism involves two main steps: the addition of methanol (MeOH) to a carboxylic acid to form a tetrahedral intermediate, and the loss of water (H2O) to create the ester.
Explanation:The Fourier esterification is a fundamental mechanism in Chemistry and encompasses two main steps. Part 1 involves the Addition of Methanol (MeOH) to a carboxylic acid to form a tetrahedral intermediate. Here, the lone pair of electrons on the oxygen of the MeOH attacks the carbonyl carbon of the acid, pushing the electrons up onto the carbonyl oxygen and forming the tetrahedral intermediate. Part 2 is the Loss of water (H2O) to create the ester. The hydroxyl group of the intermediate acts as a leaving group, departing as a water molecule. The lone pair of electrons on the oxygen then comes back down and forms the pi bond, hence forming the ester.
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In the Fischer esterification mechanism, a carboxylic acid and an alcohol react, leading to the formation of an ester and water. It's called a 'key tetrahedral intermediate' and the process involves the loss of a water molecule.
Explanation:The Fischer esterification mechanism is a fundamental process in organic chemistry that involves the transformation of a carboxylic acid and an alcohol into an ester. This process is initiated by the addition of an alcohol, such as MeOH (methanol), to the carboxylic acid via a nucleophilic attack, forming a tetrahedral intermediate.
The carboxylic acid group of this intermediate subsequently loses a water molecule (H2O) in a dehydration process, resulting in the formation of the desired ester.
For example, when ethanol reacts with acetic acid (ethanoic acid), the ester known as ethyl acetate (methyl ethanoate) is produced. Notably, esters are recognized for their fragrant odors, deriving from various plants and their fruits, such as honey methyl phenylacetate.
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Biphenyl, C 12 H 10 , C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C 6 H 6 . C6H6. At 25 ∘ C, 25 ∘C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 10.3 g 10.3 g of biphenyl in 27.8 g 27.8 g of benzene?
Answer:
Vapor pressure of solution is 98.01 Torr
Explanation:
Colligative property. In this case, we apply the lowering vapor pressure.
ΔP = P° . Xm
P° - P' = Vapor pressure of pure solvent - Vapor pressure of solution = ΔP
Xm is the molar fraction for solute. We try to determine it:
Moles of solute / Total moles → Total moles = Solute moles + Solvent moles
10.3 g . 1mol / 154 g = 0.0669 moles of biphenyl
27.8 g . 1mol / 78 g = 2.32 moles of benzene
Total moles = 2.32 moles + 0.0669 moles = 2.3869 moles
Xm for solute = 0.0669 / 2.3869 = 0.028
Let's replace data:
100.84 Torr - P' = 100.84 Torr . 0.028
P' = - (100.84 Torr . 0.028 - 100.84Torr) = 98.01 Torr
Answer:
The vapor pressure of the solution is 84.9 torr
Explanation:
Biphenyl is a nonvolatile, nonionizing solute
Temperature = 25.0 °C
Vapor pressure = 100.84 torr
Mass of biphenyl = 10.3 grams
Mass of benzene = 27.8 grams
Molar mass biphenyl = 154.21 g/mol
Molar mass benzene = 78.11 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles biphenyl = 10.3 grams / 154.21 g/mol
Moles biphenyl = 0.0668 moles
Moles benzene = 27.8 grams / 78.11 g/mol
Moles benzene = 0.356 moles
Step 3: Calculate total moles
Total moles = 0.0668 moles + 0.356 moles
Total moles= 0.4228 moles
Step 4: Calculate mol fraction benzene
Mol fraction benzene = moles benzene / total moles
Mol fraction benzene = 0.356 moles / 0.4228 moles
Mol fraction benzene = 0.842
Step 5: Calculate vapor pressure of the solution
Psol = Xbenzene * P°benzene
⇒Psol = the vapor pressure of the solution
⇒Xbenzene = mol fraction of benzene
⇒P°benzene = the vapor pressure of pure benzene
Psol = 0.842 * 100.84 torr
Psol = 84.9 torr
The vapor pressure of the solution is 84.9 torr
In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M acetone + 5 mL 1.0 M HCl + 5 mL 0.0050 M I2 + 10 mL H2O What was the molarity of acetone in the reaction mixture ? The volume of the mixture was 25 mL, and the number of moles of acetone was found to be 0.020 moles. MA = no. moles A / V of solution in liters
Explanation:
Below is an attachment containing the solution.
Final answer:
The molarity of acetone in the reaction mixture is 0.8 M.
Explanation:
To find the molarity of acetone in the reaction mixture, we need to use the molarity formula: Molarity (M) = Number of moles (n) / Volume of solution (V) in liters. Given that the number of moles of acetone is 0.020 and the volume of the mixture is 25 mL (0.025 L), we can substitute these values into the formula to find the molarity of acetone:
Molarity of acetone (MA) = 0.020 moles / 0.025 L = 0.8 M
how many aluminum atoms are in 3.78g of aluminum?
Answer:
The molar mass of aluminum is 26.9 grams per mole. We start by converting the grams of aluminum to moles of aluminum.
m
o
l
A
l
=
3.78
g
A
l
×
1
m
o
l
A
l
26.9
g
=
0.1405
m
o
l
A
l
We use Avogadro's number to determine the number of aluminum atoms present in 0.1405 mol Al.
A
t
o
m
s
A
l
=
0.1405
m
o
l
A
l
×
6.022
×
10
23
1
m
o
l
A
l
=
8.46
×
10
22
a
t
o
m
s
A
l
Answer:
0.846×10²³
Explanation:
molar mass of aluminum is 26.9 g/mol.
converting the grams of aluminum to moles of aluminum.
mol Al = 3.78 g ×1 mol Al÷26.9 g = 0.1405 mol of Al
to determine the number of aluminum atoms present in 0.1405 mol Al we use the Avogrado's number which is equal to 6.022×10²³
Atoms of Al = 0.1405 mol Al×6.022×10²³
= 0.846×10²³
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At the end of the isomerization reaction, what chemical is used to quench the residual bromine?At the end of the isomerization reaction, what chemical is used to quench the residual bromine?
Answer: Cyclohexene
Explanation:
Cyclohexane belongs to the Alkenes family. Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom get attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. When bromine is added to cyclohexane in the dark room, there won't be any reaction. If the mixture is exposed to light however, free bromine radicals are generated. In this condition, polybrominated products can be produced as well.
The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3).
Explanation:The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3). These compounds react with bromine to form non-toxic salts that can be easily removed. For example, NaHSO3 reacts with bromine to form sodium bromide (NaBr) and sodium sulfate (Na2SO4). This reaction effectively removes the residual bromine and prevents it from causing further reactions or harming the desired product.
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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.200 M benzoic acid and 25 mL of 2.00 M NaOH? Look up Ka values on the formula sheet.
Explanation:
The given data is as follows.
[HCOOH] = 0.2 M, [NaOH] = 2.0 M,
V = 500 ml, [Benzoic acid] = 0.2 M
First, we will calculate the number of moles of benzoic acid as follows.
No. of moles of benzoic acid = Molarity × Volume
= [tex]2 \times 0.475[/tex]
= 0.095 mol
And, moles of NaOH present in the solution will be as follows.
No. of moles of NaOH = Molarity × Volume
= [tex]2 \times 0.025[/tex]
= 0.05 mol
Hence, the ICE table for the chemical equation will be as follows.
[tex]C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O[/tex]
Initial: 0.095 0.05 0 0
Equlbm: (0.095 - 0.05) 0 0.05
pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]
= [tex]4.2 + log \frac{0.05}{0.045}[/tex]
= 4.245
For,
[tex]HCOOH + NaOH \rightarrow HCOONa + H_{2}O[/tex]
Initial: 0.2x 2(0.5 - x) 0
Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)
As,
pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]
4.245 = 3.75 + [tex]log \frac{Base}{Acid}[/tex]
[tex]log \frac{Base}{Acid}[/tex] = 0.5
[tex]\frac{Base}{Acid}[/tex] = 3.162
Now,
[tex]\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}[/tex] = 3.162
x = 0.464 L
Volume of NaOH = (0.5 - 0.464) L
= 0.036 L
= 36 ml (as 1 L = 1000 mL)
And, volume of formic acid is 464 mL.
36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.
We can arrive at this answer through the following calculation:
Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.Amount of moles of NaOH: [tex]2 * 0.025 = 0.05 mol[/tex]
Amount of moles of benzoic acid: [tex]2*0.475=0.095mol[/tex]
In this case, we can calculate the pH produced by the buffer of these two reagents, as follows:[tex]pH=pK{a}+log\frac{base}{acid}[/tex]
[tex]4.2+log\frac{0.05}{0.045}=4.245[/tex]
We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows:
[tex]pH=pK{a} +log\frac{base}{acid} \\4.245=3.75+log\frac{base}{acid} \\log\frac{base}{acid}=0.5\\\frac{base}{acid} = 3.162[/tex]
Now we must solve the equation above. This will be done using the following values:[tex]\frac{2(0.5-x)}{0.2x-2(0.5-x)} =0.464L[/tex]
With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.NaOH volume:
[tex](0.5-0.464) L\\0.036L ----- 36mL[/tex]
HCOOH volume:
[tex]500 mL-36mL = 464 mL[/tex]
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A magnesium hydroxide solution is prepared by adding 10.00 g of magnesium hydroxide to a volumetric flask and bringing the final volume to 1.00 L by adding water buffered at a pH of 7.0. What is the concen- tration of magnesium in this solution? (Assume that the temperature is 25◦C and the ionic strength is negligible).
Answer:
0.1724 M is the concentration of magnesium ions in this solution.
Explanation:
Mass of magnesium hydroxide gas = 10.00 g
Molar mass of magnesium hydroxide = 58 g/mol
Moles of magnesium hydroxide = [tex]\frac{10.00 g}{58 g/mol}=0.1724 mol[/tex]
Volume of the solution = V = 1.00 L
Molarity or concentration of the magnesium hydroxide:M
[tex]M=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex]
[tex]M=\frac{0.1724 mol}{1.00 L}=0.1724 M[/tex]
1 mole of magnesium hydroxide contains 1 mole of magnesium ion.Then 0.1724 M of magnesium hydroxide will have :
[tex][Mg^{2+}]=1\times 0.1724 M=0.1724 M[/tex]
0.1724 M is the concentration of magnesium ions in this solution.
The concentration of magnesium hydroxide in the solution is 0.1714 Molar. This is calculated by dividing the number of moles of magnesium hydroxide by the volume of the solution in the volumetric flask.
Explanation:The subject of this question falls under Chemistry, and it's asking about calculating concentration, specifically for a solution of magnesium hydroxide. In chemistry, concentration is commonly expressed in moles per liter (M). It's necessary to convert the mass of magnesium hydroxide to moles by using its molar mass.
First, to solve this question, we need to know the molar mass of magnesium hydroxide (Mg(OH)2), which is about 58.3197 grams/mole. Dividing the given mass of magnesium hydroxide (10.00g) by the molar mass gives us the number of moles. Thus, 10.00 g / 58.3197 g/mol ≈ 0.1714 mol of Mg(OH)2.
The whole solution was made up in a 1.00 L volumetric flask. Concentration is defined as the amount of solute per unit volume of solvent. Hence, the concentration of Mg(OH)2 is 0.1714 mol / 1.00 L = 0.1714 M (molar).
On note, the temperature given in the question (25◦C) does not affect the calculation of the concentration in this case, nor does the pH of the water used.
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Where would you expect to find the 1H NMR signal of (CH3)2Mg relative to the TMS signal? (Hint: Magnesium is less electronegative than silicon.)
The methyl protons of (CH3)2Mg are in a more electron rich environment than the methyl protons of TMS. Thus the protons of (CH3)2Mg show a signal upfield from TMS.
Explanation:
Mg is less electronegative than Si. The methyl protons of (CH3)2Mg are in a more electron-rich environment than the methyl protons of TMS.This electron density shields the protons of (CH3)2Mg from the applied magnetic field. Therefore, they sense a smaller effective magnetic field than TMS. The effective magnetic field is directly proportional to the frequency.Thus the protons of (CH3)2Mg show a signal at a lower frequency than TMS. In other words, protons of (CH3)2Mg show signal upfield from TMS.Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass of sodium reacted? Express your answer to three significant figures and include the appropriate units.
Answer: 368 grams of sodium reacted.
Explanation:
The balanced reaction is :
[tex]2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)[/tex]
[tex]\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles[/tex]
[tex]\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol[/tex]
According to stoichiometry :
2 moles of [tex]NaCl[/tex] are formed from = 2 moles of [tex]Na[/tex]
Thus 16 moles of [tex]NaCl[/tex] are formed from=[tex]\frac{2}{2}\times 16=16moles[/tex] of [tex]Na[/tex]
Mass of [tex]Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g[/tex]
Thus 368 grams of sodium reacted.
A student performs an experiment similar to what you will be doing in lab, except using titanium metal instead of magnesium metal. The student weights out 0.108 g of titanium. How many moles of titanium is this?
Answer:
n = 0.0022 mol
Explanation:
Moles is denoted by given mass divided by the molar mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molar mass .
From the information of the question ,
w = 0.108 g
As we known ,
The molar mass of titanium = 47.867 g / mol
The mole of titanium can be caused by using the above relation , i.e. ,
n = w / m
n = 0.108 g / 47.867 g / mol
n = 0.0022 mol
Final answer:
To find the number of moles of titanium in 0.108 g, divide the mass by the molar mass of titanium (47.867 g/mol), which yields approximately 0.002256 moles of titanium.
Explanation:
Calculating Moles of Titanium
The student asks: How many moles of titanium is 0.108 g? To answer this, we first need the molar mass of titanium, which is approximately 47.867 g/mol. Using the formula for calculating moles:
Number of moles = Mass (g) / Molar mass (g/mol)
So for titanium:
Number of moles of Ti = 0.108 g / 47.867 g/mol
After performing the division:
Number of moles of Ti = 0.002256 moles
This result signifies that 0.108 g of titanium corresponds to approximately 0.002256 moles of titanium.
Based upon the Aggie Honor System Rules and the Academic Integrity statement on the CHEM 111/112/117 syllabus, assess whether the statements are true or false.
Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation.
An oral discussion with a classmate regarding a paper's topics and format is cheating.
Purchasing papers from tutoring companies is allowed.
Sending a paper that you wrote to a student who is currently in another section is an honor violation.
Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data.
Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation.
Answer:
Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation. True
An oral discussion with a classmate regarding a paper's topics and format is cheating. False
Purchasing papers from tutoring companies is allowed. False
Sending a paper that you wrote to a student who is currently in another section is an honor violation. True
Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data. True
Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation. True
Explanation:
The Aggie code of honour is a code of academic integrity of the Texas A&M University. It spells out the codes of academic integrity and responsible research. Instructors are to include this code in all syllabi. It targets the application of the highest degree of integrity in academic research and forbids misconducts such as cheating, plagiarism, falsification et cetera.
The statements regarding the Aggie Honor System and Academic Integrity rules for CHEM 111/112/117 are generally correct, with exceptions for discussions about paper topics, which is not innately cheating, and purchasing of papers, which is not allowed.
Explanation:Based on the Aggie Honor System Rules and the Academic Integrity statement on the CHEM 111/112/117 syllabus, the following assessments can be made about the statements:
True: Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation.False: An oral discussion with a classmate regarding a paper's topics and format is not necessarily cheating, unless specific information about the paper is disclosed.False: Purchasing papers from tutoring companies is not allowed under the honor code's stipulations against plagiarism.True: Sending a paper that you wrote to a student who is currently in another section is an honor violation.True: Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data.True: Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation.Learn more about Academic Integrity here:https://brainly.com/question/32196816
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