New answers tagged

-2

Whatever the nature of the solute, there is no formula for calculating the density of the solution obtained by dissolving any amount of this solution into water.


2

I don't have specific product recommendations (and I think they're discouraged in general on SE), but these are the general principles: You'll only get condensation on goggles (or anything else) if they're cooler than the dewpoint of the air contacting them. If your goggles are cold, you'll get condensation. If you're sweaty, it'll be worse. If the ...


0

Most goggles with side shields should have some ventilation source on them. I'm assuming your goggles look somewhat like this: The vents on the side should facilitate enough ventilation and prevent defogging. If you're using a full face goggle with a shield, It should look something like this: The black things on the side are covered vents: they prevent ...


0

Your guess of formation of Acrolein is a pretty good one, as indicated in this answer, but a boiling water bath will probably not give you the 160°C temperature you are looking for to convert glycerol to acrolein This question is most similar to yours and has a pretty good answer by bon. The answer says: However, from personal experience, gentle heating ...


4

Both method 1 and method 2 are wrong. The concept of useless significant figures should be wiped out from school curricula and replaced with proper statistics. This significant figures and concept of estimation was good for the 18th century when eye-balling and manual measurements had to be made. The concept originated in the 16th century. In method 1: If ...


-1

Here is an example of a study of fish oil nanoemulsions which examines, to quote: The influence of surfactant-to-oil-ratio on particle size and physical stability was evaluated. Here is yet another study to quote: Although the role of surfactant in the synthesis of particles in the nanometer range has frequently been documented...The surface coverage ...


-1

The Nickel(III) hydroxide is prepared by action of Bromine $Br_2$ on Nickel(II) hydroxide $Ni(OH)_2$in presence of a $KOH$ solution. The equation of the reaction and the final formula are not established for sure. It may be : $$Ni(OH)_2 + 1/2 Br_2 + KOH -> Ni(OH)_3 + KBr$$ or$$Ni(OH)_2 + 1/2 Br_2 + KOH -> NiO(OH) + KBr + H_2O$$ Now to ...


2

Aluminium iodide ($\ce{AlI3}$) is first introduced in 1984 as an easily accessible, a highly regioselective versatile ether-cleaving reagent with novel cleavage pattern (Ref.1). Application of $\ce{AlI3}$ in organic synthesis has been reviewed (Ref.2) and it seems like acetonitrile is the best solvent to be used in catechol ether demethylation (e.g., 2-...


0

What you are measuring is not the potential $Fe^{2+}/Fe^{3+}$. It is the potentiel of the couple $[Fe(CN)_6]^{4-}/[Fe(CN)_6]^{3-}$ which is equal to +0.36 V in the table.


2

Reynolds Kitchens says Relnolds Wrap® is $98.5\%$ $\ce{Al}$ with the rest mostly $\ce{Fe}$ and $\ce{Si}$ to improve strength. You know this because when you dissolve it in $\ce{NaOH}$ solution to produce $\ce{H_2}$ a black residue remains undissolved; pure $\ce{Al}$ metal would leave no residue. Perhaps $\ce{Al}$ wiring would be more nearly pure because ...


5

From the Institute of Minerals, Materials and Mining, Most aluminium foil is made from pure aluminium, but increasingly alloys are used to improve properties and reduce the thickness required. If you want to make sure that the Aluminium foil you're using is pure, the only way would probably to ask the manufacturer. Other than that, it could be the case ...


27

First, why other options are not really the options: A: vinegar, being a weak acid, doesn't neutralize sulfuric acid and only dilutes it; B: solid sodium hydroxide, a strong base, does neutralize sulfuric acid, but it does so vigorously releasing substantial amount of heat per unit of time: $$\ce{2 NaOH(s) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)}$$ Using ...


1

Your analysis of the expectation is correct, and we can look at it more quantitatively in this way: Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of: $$-\frac{d[\...


-1

All the reaction have different concerntration at start . As you have taken so much excess NaOH it would remain constant. We could look it as you have a solution of 50 ml with 0.05 mM violet red and 0.5M NaOH . You dilute the N times and then to have to compare rates. Threotically we could find that but when we say discolouration in the solution does it mean ...


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