New answers tagged aqueous-solution
2
First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:
$$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$
$$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$
Your saying that reaction \eqref{rxn:2} might ...
3
I believe that evaporation would work, but for technical reasons, I would evaporate it to below half the original volume, cool it, and then add distilled water until the volume is half of the original level.
4
The best approach would be to crystallize $\ce{ZnSO4.7H2O}$ and then weigh the required amount and prepare a fresh solution.
How:
Slowly evaporate the solution until it is about one-fifth of its original volume. Caution: Do not boil the solution as it may spit.
Allow the concentrated solution to cool until crystals form.
Filter off the crystals and put the ...
1
2 mm drops of any liquid in any other liquid will be very unstable, and will grow to be an entire layer. Perhaps you could buy some plastic spheres of the size you desire, made from plastic slightly heavier than water. Then you can adjust the rate of fall by thickening the water. If you use an organic thickener, like xanthan gum, you should probably add a ...
1
You may ease the droplets of water to "fall further down" across the oil by adding some of dishwasher into the water.
As just tested with a glass filled with some of sunflower oil, the water-dishwater droplets reaching the surface air./.oil will deform and flatten when they fall from about 30 cm above on said surface. Eventually however, cohesion and ...
2
Electrostatic screening associated with solvation is obviously the key factor enabling dissociation in aqueous solution when compared to the gas phase. Without such dielectric screening, dissociation would be too costly ($\approx \pu{1 MJ/mol}$ or $\approx\pu{14 eV}$, roughly the energy required to ionize a hydrogen atom) to occur to a significant extent at ...
8
Magnesium perchlorate is far from unique. In fact, if you're willing to spend a little money at that hardware store you could pick up some calcium chloride, whose eutectic reaches about -50°C, not quite as low as magnesium perchlorate but still good enough to cover much of the temperature range on Mars.
Hydrogen chloride, which becomes ionic upon ...
18
Your Question: Which "exotic salt" can lower water's freezing point by $\pu{-70 ^\circ C}$?
Here is your "exotic compound" although it is not a salt by definition. It is a base: Aqua ammonia, also called ammoniacal liquor, ammonia liquor, or ammonia water, is produced by dissolving ammonia gas ($\ce{NH3}$) in water. The proper chemical name of aqua ammonia ...
44
I recently got a chance to attend a talk by someone who was working on developing analytical instrumentation on Mars. The interesting story is that the initial results by ion-selective electrode was that Mars soil is full of nitrates. Nobody knew on Earth that the nitrate ion selective electrode is far more responsive to perchlorate than nitrate. After ...
1
As the comment above suggests, if there was an obvious way to do this it would already be making someone wealthy.
Self heating cans (and gloves) are seen where an exothermic chemical reaction or heat of solution is used to thermally warm food or gloves. The most common application of this method are 'flameless ration heaters' used to heat MRE (military ...
1
The parallel thoughts:
The source water I am working with is:
12.6 mg/L or 0.126 mmol/L GH as CaCO3 equivalent
11.3 mg/L or 0.113 mmol/L KH as CaCO3 equivalent
4.8 mg/L or 0.12 mmol/L Ca
0.2 mg/L or 0.0082 mmol/L Mg
1.5 mg/L or 0.065 mmol/L Na
And the desired target water is:
70 mg/L or 0.7 mmol/L GH
40 mg/L or 0.4 mmol/L KH
20 mg/L or 0.5 ...
2
You don't measure $\Delta G^\circ$. You measure $K$ and then compute $\Delta G^\circ = -RT\ln(K)$. Crudely put, the value of $K$ will depend on the units you use for concentration and partical pressures so we have decided to use 1 M and 1 barr, respectively, when computing $\Delta G^\circ$.
Furthermore, we assume that the gases and solutions are ideal (i.e....
3
The standard states are limiting states. In these limits you can write the chemical potential of the component as
$\mu = \mu^\circ+RT\log(x)$
where $x$ is a measure of the concentration of solute (for Henry's law, where the standard state is approached as $x \rightarrow 0$) or solvent (for Raoult's law, where the standard state is approached as $x \...
0
Can I use the Henderson–Hasselbalch equation on reactions that are not buffers?
Yes, but you should still keep in mind some limitations. The equation you provide
$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$
should more generally be written
$K_\mathrm{a} = \frac{a_\ce{A-}a_\ce{H3O+}}{a_\ce{HA}a_\ce{H_2O}} $
It reduces to your equation$^\...
0
Can I use the Henderson-Hasselbalch equation on reactions that are not buffers?
If you know the equilibrium concentrations of weak acid and conjugate base, you can use the equation. You can also use it when you know the pH and the pKa and want to know the ratio of protonated and deprotonated species (for example for relating color and pH when working with ...
3
You are right with copper does not react with aluminium ions.
But as a side reaction, copper may get slowly oxidized by oxygen and dissolve in mildly acidic solution of aluminium salt.
Such a thing may happen, if you decalcify copper heating spiral by vinegar and let it stay overnight.
$$\ce{ 2 Cu + 4 CH3COOH + O2 -> 2 (CH3COO)2Cu + 2H2O }$$
...
1
Yes, the reaction
$$\ce{2 Na2SO3 + 6 SnCl2 + 12 HCl → 4 NaCl + 5 SnCl4 + SnS2 + 6 H2O}$$
is described in chapter 35.6.4 Reactions of chalcogen oxo-compounds of the Russian translation [1, p. 526] of Blumenthal's Anorganikum 10th Ed. (in German).
Unfortunately, I don't posses original German edition, so the following is my English translation of the ...
2
Or does some reaction happen that I'm not aware of?
There are three reactions, and you are probably aware of all of them:
$$\tag{1}\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
$$\tag{2}\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$
$$\tag{3}\ce{Ch3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$
The concentration of $\ce{H+}$ in those three reactions is the same because ...
5
The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.
In your answer, you are missing a negative sign in $\Delta H=−Q$ the way you start out with $Q$ from the ...
3
As a quicker way, you could also use the Pearson's square to solve this:
1.0 M - 0.4 M = 0.6 M
2.5 M - 1.0 M = 1.5 M
The ratio is 0.6 to 1.5 (which corresponds to 1 to 2.5). Notice that we did not use the final volume in this calculation, the answer does not depend on it.
0
But my question is: Does adding more solute (like adding more salt to water) increase the rate of solvation since the contact surface area is larger?
Not really. If I get the point, you would like to know if x + y grams of a substance dissolve faster (as dn/dt) than an amount x. No. Ignoring activity of the solute, or in sufficiently diluted solution, the ...
1
If the answers are to be exclusive, then obvious choice is agitating the mixture.
If the answers are not to be exclusive, then agitating the mixture + increased surface by adding more solute increases $\frac{\mathrm{d}c}{\mathrm{d}t}$
Adding more solute does not increase rate of solvation, but as the absolute surface increases, it increases $\frac{\...
3
In addition to the equation for the preservation of the amount of substance you've written, there is one more equation you can use for the system, namely the one reflecting preservation of volumes: $x + y = 3$:
$$
\left\{
\begin{align}
x + y &= 3 \\
2.5x + 0.4y &= 3
\end{align}
\right.
$$
$$x + y = 2.5x + 0.4y$$
$$1.5x = 0.6y$$
$$\frac{x}{y} =...
0
$\ce{SO4^2-}$ is actually going to make the pH basic because $\ce{HSO4-}$ is a weak acid (not $\ce{H2SO4}$, only the first hydrogen ionizes completely). By itself aqueous $\ce{Mg^2+}$ will not form magnesium hydroxide in water. But with the $\ce{OH-}$ from the $\ce{HSO4-}$ is will form a suspension with some dissolved and some not so dissolved, but still the ...
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