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 ...


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 ...


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. ...


4

You are not going to dissolve 304 and leave the concrete. Use sandpaper to remove the surface rust as needed : Or mechanically ( cut/grind) the 304 to remove it.


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 ...


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


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 (...


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 ...


3

How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions? There are three equations: $$\ce{HCOOH <=> H+ + HCOO-}$$ $$\ce{CH3COOH <=> H+ + CH3COO-}$$ $$\ce{H2O <=> H+ + OH-}$$ Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ...


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+ <=> ...


2

The conjugate base of 2-furoic acid is aromatic, but so is the acid. ' The aromaticity does not contribute to the stability of the conjugate base because the negative charge from the carboxylate anion cannot delocalize into the ring. The major resonance contributors for the carboxylate anion are: However, because of the oxygen atom, furan is an electron-...


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 ...


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 ...


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 ...


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

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, ...


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....


1

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


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 ...


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 ...


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 ...


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/...


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, ...


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