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Solution
By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$:
$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$
The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$:
$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$
The amount of substance $n$ can be found using mass of solute $m$ and ...
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In agriculture, potassium silicate ($\ce{K2SiO3}$) is used as foliar fertilizer ($\ce{K2SiO3}$ is water soluble). The major synthetic method of the preparation of $\ce{K2SiO3}$ is heating $\ce{SiO2}$ and $\ce{K2CO3}$ at high temperatures $(\pu{600-850 ^\circ C})$ in various mole ratios (e.g., Ref.1):
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2} \tag1$$
...
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The fact that $\ce{CO2}$ does not have a dipole moment does not really mean it's nonpolar. What $\ce{CO2}$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).
Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $\ce{CO2}$ the component dipoles ...
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Yes it does.
Some years ago I have made an experiment to extract potash K2CO3 from oak ashes.
I have evaporated the filtered leach liquid in an Pyrex dish while having the liquid close to the boiling point.
This whole operation made the Pyrex dish milky looking – respectively the K2CO3 did attack the glass.
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Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging.
I reference the attached article from the International Journal of Medical Sciences from 2013:
Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-...
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Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:
Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. $Na2CO3$ has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose ...
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The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance.
What can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite ...
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Assume concentration of solution A and B are $a\%$ and $b\%$, respectively (both are $\%(w/v)$). Suppose $a\% \approx b\% \approx 5\%$. Molar mass of ammonia ($\ce{NH3}$) is $\pu{17.03 gmol-1}$. Thus, if molarity of A and B are $M_A$ and $M_B$, respectively, then:
$$M_A = \frac{a \ \pu{g} \ \ce{NH3}}{\pu{100 mL}} \times \frac{\pu{1 mol}}{\pu{17.03 g} \ \ce{...
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For housewives, a solution is simply to add Epsom salt in equal doses to each mix and let the $\ce{Mg(OH)2}$ settle out of solution.
Decant and repeat the experiment until one solution no longer produces any results of a white precipitate. The latter brand is weaker in its aqueous ammonia concentration based on the reaction:
$$\ce{\overset{Epsom salt}{MgSO4 ·...
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