New answers tagged aqueous-solution
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I guess You can do electrolysis of sodium alluminate.
The result is Al and Na ions.
But the aluminum formed, converts to aluminum hydroxide and aluminum oxide very fast.
And some of the aluminum compounds can react with NaOH again. And this can creat a loop and waist energy. But for your application, I think you can do a trick to minimize waist.
Add sodium ...
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According to IUPAC, the official name for molarity is amount of substance concentration, and its symbol is $c$. The symbol for molality is $m$ or $b$, and the symbol for molar mass is $M$.
what are the meaningful differences between molarity and molality?
Molality does not change with temperature because mass and chemical amount is temperature-independent. ...
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Effect of side reactions on galvanic cell using iron rod as anode and sodium chloride as salt bridge
Since you state, "We measured a voltage way below the expected value and it was dropping quite fast," it indicates a few things that could lead to those observations.
How did you measure voltage? If you used a low impedance voltmeter, it would show much lower voltage because the impedance of the meter is low compared to that of the cell. Voltage ...
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Phenolphthalein is colorless for acidic solution, viz. upto $\mathrm{p}H=8$, and for $\mathrm{p}H>8$, it shows pink color. $\ce{NaHCO3}$ is an amphoteric salt with $\mathrm{p}H$ value close to 8, so phenolphthalein will be colorless. But $\ce{Na2CO3}$ being a basic salt, solution will show pink color.
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Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed.
Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has ...
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Here is a passage from Marschner P. (2012). Marschner’s mineral nutrition of higher plants. 3rd Edn. London: Academic Press:
In the context of the wikipedia table, it means that while you can have a high concentration of sulfate, you should consider the cation of the substance (e.g. calcium sulfate, potassium sulfate, ammonium sulfate) as well when trying ...
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This can be explained by Le chaterlier principle :
" Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium" .
When you add pure ester the reaction will try to undo the effect of change in concentration of ester ...
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Determining the crystal lattice energy would be helpful in comparing the starting points for a solubility investigation. Perhaps an approximation to the crystal stability is provided by the melting points: for fumaric acid, the mp is 287$^o$C; for maleic acid, 135$^o$C. 135$^o$C is a sort of normal mp for organic compounds. 287$^o$ seems quite high, ...
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The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does ...
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The glycoside amygdalin is a product of enzymatic biosynthesis. Its hydrolysis produces 2 molecules of glucose, benzaldehyde and hydrogen cyanide. Mixing potassium cyanide and sugar ( = sucrose = combined glucose + fructose) does not produce amygdalin. Neither you would get it, mixing glucose, benzaldehyde and hydrogen cyanide.
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From Solid-Liquid Phase Diagram of the System Methanol-Water by G. A. Miller and D. K. Carpenter in J. Chem. Eng. Data 1964, 9, 3, 371–373 (July 1, 1964) [https://doi.org/10.1021/je60022a017]
"A solution of methanol and water that is methanol-rich tends to form
a glass when cooled below the melting point. The liquid is very
viscous at such temperatures ...
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I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that ...
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Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:
$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\...
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Notice, $K_\text{f}$ for water is $1.86 \pu{°C⋅kg⋅mol^{-1}}$. This is just a constant to know. Now that you know $K_f$, you can solve for $\Delta T$. Assume 1 litre of solvent. That means there will be 7.71 moles of LiCl. Next, the mass of 1L of solution is $1.1 678\pu{cm^{-1}}$ because $\pu{g/cm^3}$ is is the same as $\pu{kg/L}$. Next, you can subtract the ...
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You probably get the $\mathrm{LD_{50}} = \pu{1585 mg/kg}$ value by reading the abstract of the paper. That's why you got confused by that value. However, it is not $\pu{1585 mg}$ of fresh leaves. It is actually $\pu{1585 mg}$ of dry leave extract. I mean dry residue obtained by removal of water from an aqueous extract of dried leaves. The authors have made a ...
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At the core of your question there is misunderstanding of what the papers you refer to are discussing. The LD per se in not related to the extract concentration nor to the naturally occurring principle(s) in the plants.
You seem interested in a relation between the original concentration and the extracts concentration. This will depend on several factors ...
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They have to analyse extracts for the plant drug concentration. They have idea about typical drug content in plant and use accordingly the ratio between solvent volume and plant leaves mass, together with various concentration techniques like vaporization or solvent/solvent extraction.
The drug concentration in final solution is then multiplied by the used ...
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As @Poutnik has already very well described the intuitive way to solve such problem so I will try to describe a standard method for such problems.
First analyse the reaction and find out the species which are being oxidised and reduced by finding the change in their oxidation state. Initially ignore all other ions which are not included in the redox reaction....
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The final arbiter of formal correctness of chemical reaction enumeration are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That ...
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The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions.
Thus the unbalanced reaction with correct oxidation states would be:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(...
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More ions form pairs when concentration of acid increases, does that decrease the speed of reaction?
In general, the conductivity of any liquid increases with the concentrations of its ions, and decreases with their dimensions. In pure sulfuric acid, the only ions existing are $\ce{H3SO4+}$ and $\ce{HSO4-}$. They are big and their concentration is low. So the conductivity of pure sulfuric acid is low. But when you add some water, the $\ce{H2O}$ molecules ...
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