9

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\...


8

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 ...


6

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 ...


4

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 ...


4

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....


3

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}(...


3

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 ...


3

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.


1

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.


1

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 ...


1

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, ...


1

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 ...


1

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 ...


1

You can make the silicates by fusing the hydroxides with silica. Potassium silicate fertilizer grade are successfully produced by direct fusion of silica ($\ce{SiO2}$) and potasium compounds($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of mixture. The fusion temperature reaches around 1350 °C. $$\ce{SiO2 + 2 KOH ->[\Delta] ...


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