New answers tagged

3

No answer is correct, but "A" is closest to the correct answer. Heavier molecules diffuse slower than lighter ones because the same amount of thermal energy has to move more mass. However, this relation (diffusion rate is inversely proportional to the square root of molecular weight) actually leads to 44 seconds not 64. The professor who told you "D" ...


2

The reaction of chloride ion with silver ion is 1:1 and highly quantitative. It is one of the best studied reactions inorganic chemistry. You cannot carry this reaction in basic/alkaline medium by any means. There is no need of any catalyst. This was worked out at least 120 years ago by people who were interested in atomic weight determinations. Theodore ...


2

If you dissolve silver nitrate $\ce{AgNO_3}$in water, and add some NaOH solution to make it basic, it will destroy your substance, and produce a brown precipitate of silver oxide $\ce{Ag_2O}$ : $$\ce{2 AgNO_3 + 2 NaOH -> Ag_2O + 2 NaNO_3 + H_2O}$$ But instead of using a hydroxide like $\ce{NaOH}$, you may add some ammonia solution $\ce{NH_3}$. In this ...


-2

I have used this complex as the hydroxide to react with Mg(NO3)2. This resulted in a precipitate of white Mg(OH)2 and the tetra-amine cupric nitrate: $\ce{[Cu(NH3)4(H2O)2](OH)_2}$ + Mg(NO3)2 -> $\ce{[Cu(NH3)4(H2O)2](NO3)_2}$ + Mg(OH)2 (s) After separating out the Mg(OH)2, I let the solution slowly evaporate which resulted in a non-energetic hydrate. With ...


4

Nowadays you have to differentiate between pH measurements with non-glass sensors and with glass sensor. In the typical pH meter with a glass sensor, there could be many things that could go wrong resulting in biased results. In general, some electrodes may be sensitive to interfering ions such as $\ce{Ag+}$. So some potential could be generated also by ...


0

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers. We know, that 3° carbocation is a more stable intermediate than an allyl ...


0

The stability of an acid is judged by the strength of it's conjugate base. If the conjugate base is stable, it implies that it is a weak base and hence the corresponding acid is stronger. Now consider H3PO4. The structure has one P=O bond and 3 P-OH bonds. The conjugate base would be one Hydrogen short from an OH group. Similarly with H3AsO4. However, the ...


-1

[EDIT] I start with the following definition: Dissociation, in chemistry, the breaking up of a compound into simpler constituents that are usually capable of recombining under other conditions. In electrolytic, or ionic, dissociation, the addition of a solvent or of energy in the form of heat causes molecules or crystals of the substance to break up into ...


2

Here are four examples of substances dissolving: $$\ce{C6H12O6(s) <=> C6H12O6(aq)}\tag{1}$$ $$\ce{CH3COOH(l)<=> CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)}\tag{2}$$ $$\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)}\tag{3}$$ $$\ce{SO3(g) <=> SO3(aq);\ \ SO3(aq) + H2O(l) <=> HSO4-(aq) + H+(aq)}\tag{4}$$ Because NaCl is an ionic solid, ...


3

This is a simple logical and semantic problem, which is not a problem at all. Your professor is right and wrong- both at the same time. He is creating a classification which does not exist and which is meaningless. Look at the word origin of electrolyte: Etymology from OED: < electro- comb. form + ancient Greek λυτός that may be dissolved, soluble (...


0

Ok, let's beat this to death with ICE tables. (a) Calculate the pH of a buffer system that contains 0.40 M of $\ce{NH3(aq)}$ and 0.50 M of $\ce{NH4Cl(aq)}$ . Note that the $K_\beta$ value of $\ce{NH3(aq)}$ is $1.8\times10^{−5}$. Some observations: We want the pH, not the pOH. For $\ce{NH4^+}$, $K_\alpha = \dfrac{K_\mathrm{w}}{K_\beta} = \dfrac{1.00\...


1

Your first result is OK. But : First, it does not make any sense to give a result with $\ce{10}$ significant figures when the initial data has only $\ce{2}$ significant figures. When a data is given like here $\ce{1.8 x 10^{-5}}$, it means that the author cannot be more precise, and may admit that the exact value of $\ce{K_b}$ is somewhere between $\ce{1....


4

I admit Karsten Theis has given en excellent answer for OP's question. However, I'd like to point out that this could be solve without getting confused by $\mathrm{p}K_\mathrm{b}$, which is common with novices when using the Henderson–Hasselbalch equation for buffers. The equation is derived by dissociation of weak acid ($\ce{HA}$): $$\ce{HA + H2O <=> ...


4

Buffer equation The Henderson equation for buffers is: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$ $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is: $$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{A-}]}{[...


1

HCL is a stronger acid so it will displace the bitartarate forming choline hydrochloride and tartaric acid


3

Naturally occuring amino acid with basic sidechain - L-Arginine: Oral supplementation with L-arginine at doses up to 15 grams daily are generally well tolerated. source here


2

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect. But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation . In ...


0

.Hey, Araj! Welcome to ChemSE! Thanks for posting your question here. As far as the acidity of water, It is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\...


0

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent. As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.


5

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.) In order for the resonance to stabilise a negative ...


1

Okay, so if we talk about the acidic strength of a compound, consider adding a base to a solution of all the above. Practically, a reaction is bound to take place-if possible-due to thermodynamic reasons. Hence, reaction in IV will take place since the base will try to find the most acidic Hydrogen, which is why one should consider the removal of H atom from ...


3

You can take both factors (1) and (2) combined as the reasoning for the greater acidity of the former. When you are comparing stability (in thermodynamic terms) of two conjugate bases, you are basically trying to position them with respect to each other on the energy axis of the energy profile (or reaction coordinate diagram). In this case, the reaction ...


1

So, there are many factors that affect acidity or basicity of a compound, the main factors being bond electronegativities and resonance structures. However, I believe you are correct in assuming that those pKa values seem a bit off. Upon further investigation, I found that 10.8 is the pKa of the conjugate acid of ethanamine, so the pKa of the actual compound ...


3

Sodium carbonate indeed dissociates in water : $$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$ But that is just beginning, as in solutions with carbonates, bicarbonates or dissolved carbon dioxide happen multiple chemical equilibriums: ( See a lot of info at Carbonic acid on Wikipedia): Carbonate anion partially hydrolyzes in water: $$\ce{CO3^2- + H+ <=> ...


1

The reaction system can be described as: $\ce{H2O <=> H+ + OH-}$ $\ce{H+ + CO3^{2-} <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$ Note, the net of the first two reactions imply a rise in pH in the presence of carbonate. Further, with time and carbon dioxide exposure: $\ce{H2O + CO2 <=> HCO3- + H+}$ $\ce{H2CO3 <=> H2O + CO2}$ And, ...


0

1) Let us calculate first the $\ce{NH_4F}$ problem. The ion $\ce{F^-}$ reacts partially with water according to $$\ce{F^- + H_2O <=> HF + OH^-}$$ The equilibrium constant $K_b$ of this equilibrium is defined by $$K_b = \frac{[HF][OH^-]}{[F^-]}$$ The numerical value of $K_b$ is needed, but it is not tabulated. However it can be obtained from the ...


2

If you use about equimolar amounts of zinc and the acid, the reaction would about asymptotically converge to depletion of the substance with the lower molar amount. The white precipitate is a mark the solution is saturated with zinc sulphate heptahydrate. The solution becoming thickened by the suphate may further decrease the rate of zinc dissolution. You ...


2

WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts. Assuming: (1) that concentrations can be used instead of activities (bad assumption...) (2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations. (3) Since HF is a ...


4

From your comments it seems that you are looking for an approximation of a log. I wish you clarified that in the main question without mentioning calculators. It seemed you just wanted to avoid a calculator for some unknown reasons. As Poutnik states, anyone who can post here, will certainly have access to computers and hence the ability to calculate logs. ...


2

Padé Approximation for ln(1+x) provides very interesting trade off between simplicity and accuracy ( See also Wikipedia - Padé approximant ): $$P\{ \ln( 1+x ) \} = \frac{x(6+x)}{6+4x}$$ $\ln(1) = 0$, $\ln(2) = 0.7$, $e_\mathrm{max} = 0.00685$, $e_\mathrm{max, rel} \lt 1\% $, $e_\mathrm{RMS} = 0.00258$ This is already a good and fast ...


2

It depends how you write the reaction. For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)? If we call the base $\ce{B}$ and the conjugate acid $\ce{BH+}$, you could write: $$\ce{B + H3O+ <=> BH+ + H2O}$$ This ...


1

The question asks: I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride. User Mathew Mahindaratne has provided an answer, but did not answer ...


1

Your initial calculations using Henderson-Hasselbalch equation is correct: $$ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\ \mathrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\ \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} = -0.05\\ \frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89 $$ Yet, since you have only $\pu{0.10 M}$ ammonia ...


2

There will be leaks around the edge of the paper. If the size difference is not too large (a few mm at most) you can remedy the situation as follows: (1) turn on the vacuum with only the filter paper in the funnel, (2) take a spatula and run the tip of the spoon around the edge of the filter paper, creasing it into the corner between the flat bottom and the ...


2

I in fact made this mistake once while attempting to dry casein isolate from milk. If the filter paper is larger than the diameter of the büchner funnel, you will risk there being gaps on the side (or "waves" as Vinz called it in his comment), due by crumples in the paper caused by fitting the large paper into the small diameter of the funnel. The air will ...


6

I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time. It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the ...


2

Per Wikipedia on Chlorous acid to quote: The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5): $\ce{2 HClO2 → HClO + HClO3}$ Also, here are some interesting comments on the decomposition of chlorous acid in "Kinetics and Mechanism of the Decomposition of Chlorous ...


1

It is my understanding that the main decomposition product of HClO2 is ClO2. Commercial ClO2 generation systems typically incorporate a membrane used to separate the gas from the other components, such as the solids and undesirable side products. This helps create a highly pure solution of ClO2 in water. Here are some examples: https://www.dioxide.com/...


3

Using concentrations and not activities, user GRSousaJr presents the exact solution to the problem in equations 1-8. However the simplifications don't really seem appropriate. This is a chemistry problem, not a problem in numerical analysis to solve. The gist is that we can make some simplifications based on the number of significant figures and a knowledge ...


0

The chemistry is complex given the organic content of sea water and the influence of sunlight. For example, here is an article detailing the influence of sunlight on oceans containing dissolved organic matter, to quote: Solar radiation mineralizes dissolved organic matter (DOM) to dissolved inorganic carbon through photochemical reactions (DIC ...


0

Try this, apply a cheap and effective spray onto the iron surface employing hypochlorous acid (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins), or better high surface area carbon particles in the spray, serving as a cathodic zone, in this ...


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