New answers tagged water
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The best way for determining the active chlorine amount ($\ce{ClO^-}$) in a bleach is to dilute it with water, then to add Potassium iodide $\ce{KI}$, and enough HCl solution to make it acidic. It produces $\ce{I_2}$ by the equation $$\ce{ClO^- + 2 H^+ + 2 I^- -> I_2 + H_2O}$$ Then the $\ce{I_2}$ so produced is titrated by a standard solution of sodium ...
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As I'm sure you've heard in the comments, this is no easy task. However, the easiest and maybe fastest (although possibly not the cheapest) way of testing for this mystery contaminant is just a couple of wide-range tests. What I would do:
A wide-spectrum light spectroscopy test (UV through IR)
An NMR Spectroscopy test.
Compare both results to results from ...
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A rubber balloon does require some force to get filled. A mylar balloon (typically filled with helium) can be purchased empty (or can be emptied) and requires almost no force to fill; then can be squashed manually to eject the gas. They are available in many sizes.
Rubber balloons also tend to go flat or leak, so their storage utility, especially for low m....
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Making the solution in a single step is inaccurate up to impossible.
Try to find a crystal of lead nitrate with the mass close enough to $\pu{1 mg}$
Try to weight up $\pu{1 mg}$ on scales with good enough accuracy.
Try to make such $\pu{1 ppm}$ solution stable enough.
The usual thing is to create a concentrated stock solution ( e.g $\pu{1000 ppm}$ and ...
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Russian interstate standard GOST 18293-72 "Drinking water. Methods for determination of lead, zinc and silver content" (PDF in Russian) suggests to use sulfarsazene (plumbone), which forms orange-yellow complex with lead(II).
Reported method sensitivity (spectrophotometry): $\pu{0.5 μg L-1}.$
A brief review in English [1, p. 315]:
To determine lead in ...
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2NaHCO3 --> Na2CO3+H2O+CO2 at temperatures above 50 degrees Celsius: TRUE.
However, Na2CO3+H2O --> NaHCO3+NaOH is a hydrolysis reaction: it only goes to a small extent, maybe 2-5%, and on evaporating the water, it goes back to Na2CO3. Na2CO3 melts at 851C without decomposition.
So, if you take a solution of NaHCO3 (pH ~8.5) and boil it, you will generate ...
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Actually, if one adds your two reactions:
$\ce{2 NaHCO3 <=> Na2CO3 + H2O + CO2}$
$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$
Upon cancelling, to derive the reaction:
$\ce{ NaHCO3 <=> NaOH + CO2}$
which implies that heated Sodium bicarbonate is a chemically active source of Lye, if one allows the CO2 gas to escape (which answers (a) of your ...
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You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.
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Alejandro, you are right, the electrolytes will make a difference. The difference is not in terms of hydrogen production per se, as Maurice mentions in the comments, it will be in terms of energy saving. Electrolytic reactions are "exact" in the sense that if 100 electrons reach the electrode from a battery, 100 hydrogen ions will be reduced. In other words,
...
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I did find a work, 'Electrolyte Engineering toward Efficient Hydrogen Production Electrocatalysis with Oxygen-crossover Regulation under Densely Buffered Near-neutral pH Conditions', with comments of interest.
I start with some background from the introduction:
In recent decades, drastic progress in solar fuel production has occurred: photovoltaic cells ...
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ppm is not necessarily defined for 2 quantities of the same unit.
There is ppm (w/w), ppm (V/V), ppm (w/V), ppm(n/n).
Sometimes these variants use an explicit affix, like ppm (V/V)=ppmv.
Due to the ambiguity of the default interpretation, it is highly recommended not to use ppm / ppb / ppt anymore. If one has to use them and no explicit interpretation ...
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$1$ liter water weighs $1030$ g, and contains $10.3$ mg $O_2$. The mass concentration of $O_2$ in ppm is the ratio $0.0103 g/1030$ g = $10.0$ ppm.
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One can also say that any equilibrium constant $K$ is related to the change of free enthalpy ${\Delta G°}$ of the particular reaction :
$${\Delta G° = RT lnK}$$ And this ${\Delta G°}$ does not depend on any other compound present in the solution.
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The equilibrium reaction for the auto-dissociation of water is:
$$2\text{ H}_2 \text{O}(l) \leftarrow \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
The associated equilibrium constant $K_w$ is:
$$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}$$
(Strictly speaking the expression is:
$$\frac{[\text{H}_3\text{O}^+]\times [\text{OH}^-...
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It depends how you write the reaction.
For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?
If we call the base $\ce{B}$ and the conjugate acid $\ce{BH+}$, you could write:
$$\ce{B + H3O+ <=> BH+ + H2O}$$
This ...
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The simplest way of checking the nature of your water is related to the difference in density. If you dissolve $80$ g NaCl in $1$ liter ordinary water $H_2O$, you obtain a solution with a density $1.056$ g/mL. You may add a little bit of an ink to make the demonstration more visual. Now you pick up maybe 1 mL of this NaCl solution with a pipette, and let ...
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Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface:
Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle).
However, because you have a closed container, some water will evaporate, ...
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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 ...
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