Answer:
pH=pKa
pH=4.44
Explanation:
Since the titration occur between a weak acid and a strong base.
then at half -equivalence point, the pH of the solution is equals to the pKa of the weak acid.
Therefore, pH=pKa
Ka of weak acid=3.6×10^−5
To calculate the pKa of the weak acid using the express below;
pKa =- log(Ka)
pKa=−log(3.6×10−5)=4.44
From the question, the pKa of the solution is at half -equivalence point
Then,
pH=pKa
pH=4.44
The question says that the titration occurred between a weak acid and a
strong base at half-equivalence point. Then we can deduce that the pH of
the solution is equal to the pKa of the weak acid.
pH=pKa
Ka of monoprotic weak acid=3.6×10⁻⁵
The pKa of the monoprotic weak acid will be calculated by :
pKa = - log(Ka)
pKa = −log(3.6×10⁻⁵) = 4.44
Since the pKa of the solution is at half -equivalence point
pH=pKa
pH=4.44
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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.
besides the ocean where else is water on Earth? Is it always in liquid form?
Answer:
water is everywhere, its in the air and the clouds, in the earth too!
it isnt always liquid form. like when it snows or hails.
Explanation:
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.
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Iron (II) is measured in abandoned mine drainage using a phenanthroline method (Standard Methods for the examination of water and wastewater, method 3500-Fe). In 9. solution and diluted to 100.0 ml. The solution turns deep red and is measured at 510 nm. A series of standards from 0.50 -2.00 ug/ml iron(II) are prepared and their absorbance measured. The data for these solutions is given below. Using this data calculate the concentration of iron(II) in the original mine drainage samples. Fe Conc (ug/ml) Absorbance 0.00 0.50 1.000.134 1.50 2.00 0.002 0.045 0.096 0.196 Mine drainage 0.115 0.120
Answer:
Mine drainage Sample 1: [tex][Fe(II)] sample 1 = \frac{121 \mu gFe(II)}{1ml} = 121 \frac{\mu g}{ml}[/tex]
Mine drainage Sample 2: [tex][Fe(II)] sample 2 = \frac{127 \mu gFe(II)}{1ml} = 127 \frac{\mu g}{ml}[/tex]
Explanation:
Sodium hydrogen carbonate (NaHCO3) , also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid (HC1) , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HC1 through this reaction:
HC1(aq) + NaHCO3 (aq) → NaC1 (aq) + H2O (I) + CO2 (g)
The CO2 gas produced is what makes you burp after drinking the solution.
Suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 50. mL of a 0.034 M HC1 solution. What mass of NaHCO 3 would he need to ingest to neutralize this much HC1 ?
Be sure your answer has the correct number of significant digits.
Answer:
Mass of NaHCO₃ required for the neutralization = 0.143 g
Explanation:
The reaction between HCl and NaHCO₃ is given as
HCl + NaHCO₃ → NaCl + H₂O + CO₂
The fluid in the stomach of a man suffering from indigestion is considered to be 50 mL of a 0.034 M HCl solution. The mass of NaHCO₃ he would need to ingest to neutralize this much HCl.
We forst need to calculate the number of moles of HCl in 50.0 mL of 0.034 M HCl
(Number of moles) = (Conc in mol/L) × (Volume in L)
Conc in mol/L = 0.034 M
Volume in L = (50/1000) = 0.05 L
Number of moles of HCl = 0.034 × 0.05 = 0.0017 moles
From the stoichiometric balance of the reaction,
1 mole of HCl requires 1 mole of NaHCO₃
0.0017 moles of HCl will require 0.0017 moles of NaHCO₃.
So, we can then.calculate the mass of NaHCO₃ required for this neutralization.
Mass = (Number of moles) × (Molar mass)
Molar mass of NaHCO₃ = 84.007 g/mol
Mass of NaHCO₃ required for this neutralization = 0.0017 × 84.007 = 0.1428119 g = 0.143 g to 3 s.f because calculated values usually have 1 more significant figure than the given parameters for the calculations.
Hope this Helps!!!
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.
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The decomposition reaction 2 NOCl → 2 NO + Cl_2 has a rate law that is second order with respect to [NOCl], where k = 3.2 M^{-1}s^{-1} at a certain temperature. If the initial concentration of NOCl is 0.076 M, how many seconds will it take for [NOCl] to decrease to 0.042 M at this temperature? Do not enter units with your numerical answer.
For the second-order reaction 2 NOCl → 2 NO + Cl2, it will take 3.32 seconds for the concentration of NOCl to decrease from 0.076 M to 0.042 M.
Explanation:The reaction 2 NOCl → 2 NO + Cl2 is second order with respect to [NOCl], meaning the rate law can be written as rate = k[NOCl]2. Given that k = 3.2 M−1s−1, we can use the integrated rate law for second-order reactions to find the time it takes for the concentration of NOCl to change from 0.076 M to 0.042 M. The integrated rate law for second-order reactions is ¼¾ ϑ¾ = k(t-t0), with t as the time elapsed, t0 as the initial time, and ϑ and ϑ0 as the final and initial concentrations, respectively. Solving for t gives us
t = ¼¾ [ϑ0^{-1} - ϑ^{-1}] / k
Substituting the given values we get:
t = ( ¼¾ [0.076−1 - 0.042−1] ) / 3.2
t = ( ¼¾ [13.16 - 23.81] ) / 3.2
t = ( ¼¾ [-10.65] ) / 3.2
t = 3.32 seconds
In the equation KClO3 -> KCl + O2, how many moles of oxygen are produced when 3.0 mol of KClO3 decomposes completely?
Answer:
4.5
Hope this helps!
In the given chemical equation according to stoichiometry, 4.5 moles of oxygen are produced when 3.0 mole of KClO₃ decomposes completely.
What is stoichiometry?Stoichiometry is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.
Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample. It is important while making solutions and balancing chemical equations.In the given balanced chemical equation, 2 moles of KClO₃ gives 3 moles of oxygen , thus 3 mole of KClO₃ will give 3×3/2=4.5 moles.
Thus, 4.5 moles of oxygen are produced when 3.0 mole of KClO₃ decomposes completely.
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The compound most likely to be ionic is: carbon tetrachloride, potassium fluoride, carbon dioxide, or iodine monochloride?
Answer:
Potassium fluoride
Explanation:
Because potassium is strongly electropositive combined ionically with Fluorine which is highly electronegative this has the greatest tendency of been ionic
Can convection occur in both liquids and
gases? Suggest a reason for your answer
using the particle theory.
Answer:
Yes, Convection can occur in both liquids and gases
Explanation:
The Particle Theory suggests that Particles are always moving. Convection occurs in the breeze you feel, when the warm particles quickly move upwards resulting in cold air quickly sinking and creating a breeze. Convection also occurs when you boil water, the molecules of warm water quickly move to the top and cold water quickly moves to the bottom resulting in a circle the constantly keeps the water circulating, making it boil.
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.
what do you think could be happening to the sand eels
Answer:
is there like a passage to answer this question?
Explanation:
Answer:
Increasing fishing for them is thought to be causing problems for some of their natural predators, especially the auks which take them in deeper water. They are also tied as flies to catch fish. An instance of this was the RSPB report linking a population crash of seabirds in the North Sea to fishing for sand eels.
Explanation:
You have two beakers. One beaker contains 100 mL of NaOH (a strong base); the other contains 100 mL of aqueous Na3PO4 (a weak base). You test the pH of each solution. Which of the following statements is true ?
Answer:
D. If the pH of NaOH is 12, then that of Na₃PO₄ solution has to be lesser than 12.
Explanation:
In this problem, we are comparing the pH of a strong base to a weak base. A strong base is one that ionizes completely in aqueous solutions where as a weak base ionizes slightly.
The pH scale is good tool for measuring the acidity and alkalinity of various substances. It ranges from 1 - 14;
1 7 14
← →
increasing acidity increasing alkalinity
neutrality
Strong bases have their pH value close to 14 and weak bases are close to 7.
Since Na₃PO₄ is a weak base, it will have lesser pH value compared to a strong base such as NaOH
The question lack options, that are as follows"
a. The Na3PO4 has a higher pH because it has more sodium ions than NaOH. NaOH(aq- Na3PO4(aq)b. It is possible for the solutions in each beaker to have the same pHC. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be greater than 12.00.d. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be less than 12.00.The following statements are true in the given question:
d. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be less than 12.00.We know that:
NaOH is a strong base and Na3PO4 is a weak base. As the strength of OH- in 100ml NaOH and that of OH- in 100ml Na3PO4 is different. They have different pH values. As we know that pH is inversely proportional to H+ ion concentration. This means the higher the pH value lower is H+ ion concentration.Again,
PH +POH =14
From the above equation,
the higher the PH value of a solution less will be its value of POH vice versa. Hence greater is the OH- ion concentration.So,
the pH of a solution is directly proportional to OH-ion concentration.Thus, If pH NaOH = 12.00, NaOH is a strong base and Na3PO4 is a weak base. For the solution, the weak base OH-ion concentration is less. Clearly, the pH of Na3PO4 is less than 12.00.
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Problem 19.24a Using the reagents below, list in order (by letter, no period) those necessary to convert the starting material into the given product: Note: More than 1 mole of the starting material may be used. Not all spaces provided may be needed. Type "na" in any space where you have no reagent. a. heat, -CO2 b. NaOEt c. (CH3CH2)2CuLi d. CH2Cl2, PCC e. C3H7C(O)CH(C2H5)C(O)C2H5 f. CH3CH2Li g. CH3C(O)Cl, AlCl3 h. NBS, ROOR i. H2NC(O)NH2 j. HN(CH3)2 k. OH-, H2O, heat then H3O l. H3O
Answer:
Step 1: The answer is option (b) NaOEt
Step 2: The answer is option (k) OH-, H2O, heat then H3O
Step 3: The answer is option (a) heat, -CO2
Step 4: na (no reagent)
Explanation:
See the attached file for the explanation.
Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests and the reasoning illustrated in the introduction and the Week 2 Worksheet. Be sure to include equations for all the reactions which occur at each step in the scheme. Record the scheme in the data section of the lab notebook. The scheme MUST be submitted at the start of Lab Period 2. It will be reviewed by the TA at the start of the lab period, before you begin testing your scheme. Ag+, Fe 3+ Cu2+, Ni2+
Answer:
SEPARATION SCHEME FOR CATIONS
GIVEN CATIONS : [tex]Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}[/tex]
Step 1: Add [tex]6mol/dm^3[/tex] of [tex]HCl[/tex] to the mixture solution
Result : This would cause a precipitate of [tex]AgCl[/tex] to be formed
Reaction : [tex]Ag^{+} _{(aq)} + Cl^{-} _{(aq)} ---------> AgCl(ppt)[/tex]
Step 2 : Next is to remove the precipitate and add [tex]H_2S[/tex] to the remaining
solution in the presence of [tex]0.2 \ mol/dm^3[/tex] of HCl
Result : This would cause a precipitate of [tex]CuS[/tex] to be formed
Reaction : [tex]Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)[/tex]
Step 3: Next remove the precipitate then add [tex]6 \ mol/dm^3[/tex] of aqueous
[tex]NH_3 (NH_3 \cdot H_2 O)[/tex] , process the solution in a centrifuge,when the
process is done then sort out the precipitate from the solution
Now this precipitate is [tex]Fe(OH)_3[/tex] and the remaining solution
contains [tex](Ni (NH_3)_6)[/tex]
Next take out the precipitate to a different beaker and add HCl
to it this will dissolve it, then add a drop of [tex]NH_4SCN[/tex] this will
form a precipitate [tex]Fe(SCN)_{6}^{3-}[/tex] which will have the color of
blood indicating the presence of [tex]Fe^{3+}[/tex]
Reaction : [tex]F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}[/tex]
[tex]Fe (OH)_{(s)} _3 + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}[/tex]
[tex]Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}[/tex]
Now the remaining mixture contains [tex]Ni^{2+}[/tex]
Explanation:
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.
12. The Action of Phospholipases The venom of the Eastern diamondback rattler and the Indian cobra contains phospholipase A2, which catalyzes the hydrolysis of fatty acids at the C-2 position of glycerophospholipids. The phospholipid breakdown product of this reaction is lysolecithin (lecithin is phosphatidylcholine). At high concentrations, this and other lysophospholipids act as detergents, dissolving the membranes of erythrocytes and lysing the cells. Extensive hemolysis may be life- threatening. (a) All detergents are amphipathic. What are the hydrophilic and hydrophobic portions of lysolecithin
Answer:
Explanation:
Compounds derived from Phosphatidyl choline are referred to as Lysolecithins. These compounds are amphiphatic in nature and have both hydrophilic and hydrophobic regions.
Choline, glycerol, and phosphate represent the hydrophyilic portion of lysolecithins,while the fatty acid on C1 is hydrophobic in nature.
Glycerol, choline and phosphate act as a polar tail whereas the fatty acid acts as a long polar head which represents an amphipathic molecule like detergents.
A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute NaOH, step 2?
Answer:
I would expect to extract the acetic acid.
Explanation:
In the first step, since we are adding a concentrated acid, it will react with the bases present in the mixture (diethylamine and ammonia) forming salts, which are soluble in water. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.
In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) + 2Cl- (aq) ⇄ CuCl2- (aq) K = 8.7 x 104(a) Calculate the solubility of CuCl in pure water. (Ignore CuCl2- formation for part a).(b) Calculate the solubility of CuCl in 0.100 M NaCl solution.
Answer: (a) The solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) The solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex]
Initial: 0 0
Change: +x +x
Equilibm: x x
[tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
And, equilibrium expression is as follows.
[tex]K_{sp} = [Cu^{+}][Cl^{-}][/tex]
[tex]1.2 \times 10^{-6} = x \times x[/tex]
x = [tex]1.1 \times 10^{-3} M[/tex]
Hence, the solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) When NaCl is 0.1 M,
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex], [tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
[tex]Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)[/tex], [tex]K = 8.7 \times 10^{4}[/tex]
Net equation: [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
[tex]K' = K_{sp} \times K[/tex]
= 0.1044
So for, [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = [tex]\frac{CuCl_{2}}{Cl^{-}}[/tex]
0.1044 = [tex]\frac{x}{0.1 - x}[/tex]
x = [tex]9.5 \times 10^{-3} M[/tex]
Therefore, the solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Final answer:
The solubility of CuCl in pure water and in 0.100 M NaCl solution both turn out to be 1.1 x 10^-3 M when formation of CuCl2- is ignored, determined using the provided Ksp.
Explanation:
To calculate the solubility of CuCl in pure water, we can use the solubility product constant (Ksp). The dissociation of CuCl can be represented as:
CuCl(s) ⇌ Cu+(aq) + Cl-(aq)
Let the solubility of CuCl be 's' moles per liter. At equilibrium, the concentration of Cu+ and Cl- ions will both be 's' M. So, the Ksp expression for CuCl is:
Ksp = [Cu+][Cl-] = (s)(s) = s²
Giving us:
s = √(1.2 x 10^-6) = 1.1 x 10^-3 M
This is the solubility of CuCl in pure water.
For the solubility of CuCl in 0.100 M NaCl solution, we have to consider the common ion effect. In this case, Cl- is the common ion. With the presence of 0.100 M NaCl, the concentration of Cl- ions at equilibrium will be higher. However, since we ignore the formation of CuCl2-, the solubility will still be governed by the Ksp of CuCl. The calculation will remain the same as in pure water, due to the assumption of ignoring CuCl2- formation:
s = 1.1 x 10^-3 M
Therefore, the presence of NaCl does not affect the solubility under the given conditions for this specific case.
what is the molecule OH- called?
Answer:
it hydroxide ion
Hope it will help you
The molecule OH- is termed as a hydroxide ion. It comprises one hydrogen ion and one oxygen ion and holds a negative charge. Hydroxide ions are observed in water-based solutions and play a crucial role in chemical reactions such as hydrolysis; they also impact the pH levels in solutions.
Explanation:The molecule OH- is called a hydroxide ion. This comes from the fact that it consists of one hydrogen ion (H) and one oxygen ion (O). An ion is an atom or molecule that carries a charge, in this case, the charge of the hydroxide ion is -1 because it has gained one electron. Hydroxide ions are found in solutions resulting from the ionization of water.
The molecule OH- plays an important role in various chemical reactions. For example, in hydrolysis, a molecule of water disrupts a compound; the water splits into H and OH. One part of the divided compound bonds with the hydrogen atom, and the other part bonds with the hydroxide group. The presence of hydroxide ions also has a significant effect in determining the pH of a solution; the higher the concentration of hydroxide ions, the more basic or alkaline the solution is.
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Calculate the amount of heat released to convert 150.0 g of to water to ice at 0ºC.
-50100 J
-339,000 J
-627 J
-307.5 J
Answer:
There correct answer is option 1 (-50100 J)
Explanation:
Step 1: Data given
Mass of water = 150.0 grams
ΔHfus = 334 J/g
Temperature = 0°C
ΔHfus = the amount of energy needed to change phase of 1 mol water to ice
Step 2: Calculate the amount of heat required
Q = m*ΔHfus
⇒Q = the amount of heat reuired = TO BE DETERMINED
⇒m =the mass of water = 150.0 grams
⇒ΔHfus = 334 J/g
Q = 150.0 grams * 334 J/g
Q = 50100 J of heat required
Step 3: Calculate heat released
Qrequired = -Q released
Qreleased = -50100 J
There correct answer is option 1 (-50100 J)
Answer:
The answer is -50100 J
Explanation:
I just took the test as I'm in GCA as well (11th grade). My teacher walked us through each formula and gave us multiple practice questions in Jigsaw.
Honestly, all you need to be aware of is the fusion formula which is:
q = m*Hfusion
q = amount of energy
m = mass in grams
Hfusion = 334 J/g (it will not change)
Write down what we're given and what we're looking for:
q = ?
m = 150
Hfusion = 334 J/g
Set up the equation:
q = 150 * 334
Multiply 150 by 334 to get 50100.
You must keep in mind that we're finding how much heat is released which means the energy will be negative.
Add the negative sign to the number to get -50100 Joules
I don't know why that chart is there, it doesn't do anything for this equation. Disregard it for this problem.
Hope I helped! :D
g Which compounds are bases in aqueous solution according to Brønsted–Lowry theory?A chain of four carbon atoms connected by single bonds. Single bonds connect the first carbon atom to 3 hydrogen atoms. The second and third carbons are each connected to 2 hydrogen atoms by single bonds. The fourth carbon atom is double bonded to 1 oxygen atom and single bonded to 1 oxygen atom. The single bonded oxygen atom is connected to a hydrogen atom by a single bond. Both oxygen atoms have 2 lone pairs.A chain of 4 carbon atoms connected by single bonds. The first carbon atom is connected to 3 hydrogen atoms by single bonds. The second and third carbons are each connected to 2 hydrogen atoms by single bonds. The fourth carbon atom is double bonded to 1 oxygen atom with 2 lone pairs, and single bonded to 1 oxygen atom with 3 lone pairs. The oxygen atom with 3 lone pairs is marked with a minus sign.
Answer:
A chain of four carbon atoms connected by single bonds. Single bonds connect the first carbon atom to 3 hydrogen atoms. The second and third carbons are each connected to 2 hydrogen atoms by single bonds. The fourth carbon atom is double bonded to 1 oxygen atom and single bonded to 1 oxygen atom. The single bonded oxygen atom is connected to a hydrogen atom by a single bond. Both oxygen atoms have 2 lone pairs.
Explanation:
In the Brownstead-Lowry definition of a base; a base is any substance that accepts a proton. Hence, a base must have available lone pairs to accept a proton and possess a plus sign afterwards. Only the bonding situation described in the answer meets this description.
The description in the second option should be a Brownstead-Lowry acid because it will rather donate a proton.
The compounds that are bases in aqueous solution according to Bronsted-Lowry theory is ; ( A , B, C, D and E ).
A chain of four carbon atoms connected by single bonds.Single bonds connect the first carbon atom to 3 hydrogen atomsThe second and third carbons are each connected to 2 hydrogen atoms by single bonds.The fourth carbon atom is double bonded to 1 oxygen atom and single bonded to 1 oxygen atom.The single bonded oxygen atom is connected to a hydrogen atom by a single bond.Both oxygen atoms have 2 lone pairs.Bronsted-Lowry theory states that a compound which is capable of accepting protons from another compound is a base while a compound that transfers protons to other compounds is categorised as an acid. The compounds listed in the answer are capable of accepting protons because of the availability of lone pairs of protons.
Hence we can conclude that the compounds that re bases in aqueous solutions are as listed above.
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Addition of AgNO3 to aqueous solutions of the complex results in a cloudy white precipitate, presumably AgCl. You dissolve 0.1000 g of the complex in H2O and perform a precipitation titration with 0.0500 M AgNO3 as the titrant. Using an electrode that is sensitive to [Ag ], you reach the endpoint after 9.00 mL of titrant is added. How many grams of chloride ion were present in the 0.1000-g sample
Answer:
0.016 grams of chloride ion were present in the 0.1000 grams of sample.
Explanation:
According to question, 9.00 mL of titrant was added to solution with 0.1000 grams of complex to reach the end point.
Molarity of the silver nitrate solution = 0.0500 M
Volume of the silver nitrate solution = V = 9.00 mL = 0.009 L
1 mL = 1000 L
Moles of silver nitrate = n
[tex]Molarity=\frac{Moles}{Volume (L)}[/tex]
[tex]0.0500 M=\frac{n}{0.009 L}[/tex]
n = 0.00045 mol
[tex]Cl^-+AgNO_3\rightarrow AgCl+NO_3^{-}[/tex]
According to 1 mole of silver nitrate reacts with 1 mol of chloride ion, then 0.00045 moles of silver nitrate will :
[tex]\frac{1}{1}\times 0.00045 mol=0.00045 mol[/tex] of chloride ions
Mass of chloride ions :
0.00045 mol × 35.5 g/mol = 0.016 g
0.016 grams of chloride ion were present in the 0.1000 grams of sample.
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
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.
Conceptual Checkpoint 18.13 When 1,3-dinitrobenzene is treated with nitric acid and sulfuric acid at elevated temperature, the product is 1,3,5-trinitrobenzene. Draw the sigma complex for each possible pathway to compare the relative stability of each sigma complex. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.
Using sulfuric acid in the nitration of benzene generates a more reactive nitronium ion, which then reacts with benzene. The detailed mechanisms of nitration and sulfonation involve sigma complexes with resonance stabilization. An energy diagram for nitration shows intermediates and activation energy.
The importance of using sulfuric acid in the nitration of benzene by nitric acid is to activate the HNO₃, creating a more reactive electrophile, the nitronium ion (NO₂+). This ion then reacts with benzene to form nitrobenzene. In the mechanism of the sulfonation of benzene, sulfur trioxide (SO₃) acts as the electrophile, which is generated from sulfuric acid. The detailed mechanisms for both reactions involve the formation of a sigma complex with resonance forms and the restoration of aromaticity through deprotonation.
For the nitration of benzene, an energy diagram shows the activation energy and intermediates formed. The sigma complex for 1,3-dinitrobenzene undergoing further nitration to form 1,3,5-trinitrobenzene illustrates how the meta position stabilizes the complex by avoiding the creation of high-energy intermediates with adjacent positive charges, as seen in ortho and para substitution.
The equations representing the nitration and sulfonation of benzene are as follows:
Nitration: C₆H₆ + HNO₃ -> C₆H₅NO₂ + H₂OSulfonation: C₆H₆ + SO₃/H₂SO₄ -> C₆H₅SO₃H + H₂OHow 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.
An alkyne with the molecular formula C6H10 was treated with ozone followed by water to produce only one type of carboxylic acid. Draw the structure of the starting alkyne and the product of ozonolysis.
The starting alkyne is 3-hexyne, and after the ozonolysis process and treatment with water, the resulted product is hexanoic acid.
Explanation:In the given question, an alkyne with the molecular formula C6H10 is treated with ozone followed by water, to produce one type of carboxylic acid via the process of ozonolysis. The starting alkyne for this reaction probably is 3-hexyne. During ozonolysis, the alkyne triple bond is cleaved, forming two equivalent aldehydes. But these aldehydes are further oxidized into carboxylic acids when water is used as second reagent. The produced product, hence, is hexanoic acid.
Remember, in the process of ozonolysis, alkynes are oxidized and cleaved into smaller molecules. These molecules usually have carbonyl groups such as aldehydes and ketones. But in this scenario, because of the hydration step, we end up with a carboxylic acid.
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Convert 10 liter into m3.
10 litres = 0.01 cubic meter
Be sure to answer all parts.
(a) How many atoms are directly bonded to the central atom in a trigonal planar molecule?
i. two
ii. three
iii. six
iv. eight
(b) How many atoms are directly bonded to the central atom in a trigonal bipyramidal molecule?
i. three
ii. four
iii. five
iv. six
(c) How many atoms are directly bonded to the central atom in an octahedral molecule?
i. three
ii. four
iii. six
iv. eight
Answer:
a) ii
b)iii
c)iii
Explanation:
three atoms directly bonded then only it is possible to achieve trigonal planar
trigonal bipyramidal means five atoms should attach to central atom
for octahedral six atoms must directly connected to central atom