24

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.


22

Hydrogen critical temperature is $\pu{32.938 K, resp. -240.21 ^{\circ}C}$. Above this temperature, it cannot be liquified. So to answer your question, you can get as high pressure as you can produce and the container can withstand, as there is no condensation reducing the pressure. WARNING: An accidental explosive container rupture can easily cause severe ...


7

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent. All Brønsted acids can act as oxidizing agents, because the acidic proton can be reduced to hydrogen gas." The whole sentence is written in a very convoluted fashion, and it is partially wrong as well. Someone should improve this Wikipedia section. All they are saying is that strong ...


6

At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$. Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York ...


6

As others have said, hydrogen cannot be liquified above its critical temperature, which my source (Wolfram Alpha chemical database*) says is $\pu{32.97 K} = \pu{-240.18 ^\circ C}$ However, with sufficient pressure, the molecules can be squeezed together until they have a liquid-like density**, and are thus no longer considered a gas, but rather a ...


5

Be aware that $\ce{H+}$ is nothing else then a proton $^1_1\mathrm{p}$. Protons have an extreme charge density, typically $10^{15}$ higher than any other ion ( but bare electrons or atomic nuclei). Protons can exist free in vacuum or plasma, but even in gases they quickly react with many gaseous molecules, forming molecular ions, like dioxidanylium $\ce{O2 +...


4

The definition of oxidation and reduction are Oxidation is the loss of electrons or an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule. Reduction is the gain of electrons or a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule. By this definition a $\ce{H+}$ ion can gain an electron ...


4

Hydrogen does not just insert itself physically into the metal. Although only a few metals form stoichiometric hydrides with predominantly ionic character, most of them (and all those involved in hydrogen/hydride storage) form some type of bond, generally involving saturation of the hydrogen 1s orbital with ionic or polar covalent bonding. Helium simply ...


4

Hydrogen, having very small and fast moving molecules, diffuses faster and through smaller holes than air. Your balloons seem to be permeable for hydrogen. Try a different brand, if it helps. But as others have mentioned, rubber is not the best material to keep hydrogen. Better is a metallised plastic. Pure hydrogen just burns in a flame. If you have ever ...


3

You may have confused barium peroxide with barium metal. The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid. So your missing product is water, not hydrogen. Tricky half-reactions If we ...


3

Your problem stems from the statement "I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+." This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{...


3

The half-equations are not correctly written. There is no Oxygen atom released, and no H atom emitted, as the author proposes. And the cathode does not get 4 electrons, as he or she states. The correct half-equations should be, first at the anode : $$\ce{Pb + SO_4^{2-} -> PbSO4 + 2 e^-}$$ And at the cathode it is : $$\ce{PbO_2 + 4H+ + SO_4^{2-} + 2 e^- ...


3

You will probably find all this information in biochemistry textbook. To give you an idea in proteins, typically; alpha helices H..O distance is 2.06 Angstrom, and N..O, 2.99. In beta sheets (parallel) H--O 1.97, N--O 2.29 and in anti parallel sheets, H--O 1.96 and N--OI 2.91. These H bonds are not linear, e.g C=O..H-N typically 150 to 160 degree for COH or ...


2

When you dip magnesium ribbon into a copper sulfate solution, theoretically (and realistically) you do get a simple replacement reaction. Then reality sets in: you have a metallic anode (magnesium) with little cathodes (copper) all over it. At that point, the magnesium just overreacts. Well, it reacts faster. The size of the bubbles will cause some ...


2

With decreasing electrolyte concentration, either the needed voltage raises at constant current, either the passing current decreases at the constant voltage. Both happens because at lower concentration, a higher potential gradient across the electrolyte is needed to maintain the same current, as the smaller ion count requires ions to move faster for the ...


2

All acids are oxidizing. They are all able to oxidize metals $\ce{M}$ whose redox potentials are negative with respect to hydrogen, like zinc $\ce{Zn} ~(E° = - 0.76~\mathrm V$), iron $\ce{Fe}~ (E° = -0.41~\mathrm V)$, and magnesium $\ce{Mg} ~(E° = -2.37~\mathrm V)$. The reaction produces some hydrogen gas $\ce{H2}$ and the metallic cation $\ce{M^{z+}}$. But ...


2

The primary problem when switching to different fuels in appliances or engines is getting the fuel/air mix right Previous answers contain a disturbing number of misleading hypothetical reasons why hydrogen can't work. Many are just plain wrong. For example, the idea that dealing with the water produced is ludicrous given that hydrocarbon fuels produce plenty ...


2

Emission and absorption are inverse processes. So, naturally, if an atom exhibits some emission spectrum, it will also give an absorption spectrum. So you were right in assuming that. However, the processes by which these two spectra are generated are completely the opposite of each other. An atom emits a photon when an electron goes from a higher energy ...


2

The Lyman series is made of lines in the spectrum corresponding to the return of the electron from a higher level ($n = 2, 3, 4,$ etc.) to the first and deepest level ($n = 1$) . The Balmer series is made of lines in the spectrum corresponding to the return of the electron from a higher level ($n = 3, 4, 5$, etc.) to the second level ($n = 2$) The Paschen ...


1

Aluminum cannot be used easily as support for a hydrogen electrode, because it is always covered by a thin, continuous and insulating layer of aluminum oxide. As this layer is an electrical insulator, its resistance is high, and ions like $\ce{H+}$ have some difficulty in crossing it. So the measured potential of hydrogen measured on such a protected ...


1

I understand the point of your question, here is a discussion of the associated science. The process remains a classic electrochemical process where here H2O2 is a replacement for oxygen and the require $\ce{H+}$ is provided by the $\ce{HCl}$ which is also a source of chloride ions noted to promote corrosion as well as forming salts which are excellent ...


1

If we try to answer this question at a rather simple level, we could say that yes the ions can exist in a particle-free environment like vacuum. But if ions are formed in an aqueous medium, like the H+ ions in acidified water then the H+ readily combines with H2O to form hydronium ion. Which means that the H+ is not stable but since it can't combine with ...


1

Platinum(IV) chloride adopt an octahedral coordination geometry. This geometry is achieved by forming a polymer wherein half of the chloride ligands bridge between the platinum centers. So, whenever it is dissolved in some acids, say $\ce{HCl}$, the chloride bridging ligands breaks and the molecules gets added giving $\ce{H2PtCl6}$, also called ...


1

So fluorine does have the ability to hydrogen bond. However, they are usually transient and depend significantly on the species that is donating the H-bond along with some other factors (i.e. what phase the molecules are in, temperature, etc...) Fluorine is so electronegative that it holds its electrons much "tighter" then other atoms, making it a ...


1

in the sun hydrogen is in the plasma state, which makes it impossible to form bonds (because in the plasma state electrons are separated from the nucleus) not to mention the heat alone would easily break the bonds between hydrogen and oxygen if you brought water to the sun


1

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.


1

Since the question is about the exact size — and additional citations are requested — it may be worth mentioning Randell Mills' model which calculates the hydride radius analytically as $1+\sqrt{s(s+1)}$ times the Bohr radius, where $s$ is the electron spin of $\frac{1}{2}$. This works out to 99pm and represents the radius of a free hydride ion. This is ...


1

Flame Arresters are fitted to prevent the ingress of an external ignition entering the facility, not to stop a "big fireball going outside". Due to the wide ranging LEL-UEL limits explained above, it is even more essential that a flame arrester is installed.


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