New answers tagged

0

Put some $O_{2}$ and $H_{2}$ in a chamber, give it some activation energy and boom, you got some water. However, mixing a flammable gas with a strong oxidizer can make things a little interesting. Plus we would be manufacturing large amounts, making the process very dangerous (i.e. expensive equipment is needed)


0

I have to say thank you to Maurice for his suggestion to use more concentrated HCl. I took the same pieces of floppy disk that I used before and increased the acid concentration by a factor of 10 and let them sit for another week. At first I was disappointed, because there was no visible change in the plastic or the solutions. But when I took the pieces out ...


7

I was able to reach the manufacturer and ask her directly. USPXX is the "edition" of the US Pharmacopeia's "exams" that a product must pass in order to be considered pharmaceutical grade. As time goes on, these exams may get updated and improved. So a higher number USPXX means that a product passed a more recent (and likely, more ...


-1

If you replace sulfuric acid by a sulfate, the half-equations will become : $$\ce{Pb + SO_4^{2-} -> PbSO_4 + 2 e^-}$$ $$\ce{PbO2 + SO4^{2-} + 2 H2O + 2e- -> PbSO4 + 4 OH-}$$ As a consequence, the solution will loose sulfate ions which will be replaced by $\ce{OH-}$, which is unfavorable, whatever the origin of the ion sulfate, as will be shown now. ...


-1

Disagree as your selected salts are not sources of the $\ce{HSO4- ion}$, as is cited in both half-cell reactions occurring in the Lead-Acid Battery. For example, per Wikipedia, to quote: Negative plate reaction $$\ce{Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e− }$$ And further: Positive plate reaction $$\ce{PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e− → PbSO4(s) + ...


1

The question can be reformulated to: "At what T is the equation $\exp{\left(-\frac{E}{RT}\right)}=0$ true ?" There is no temperature above which $\ce{H2O2}$ decomposes and below which it does not. It is rather about an arbitrary choice of acceptable decomposition rate. It is similar to solvent evaporation. There is no temperature above which ...


0

Actually there are commercial fire extinguishers that work by displacing the oxygen in a room. I forget the name but the safety experts put spheres containing an inert gas ( not a noble gas ,not CO2) to auto automatically be released in event of fire . Hydrogen was used in these lab rooms , but I don't know if that is why these extinguishes were chosen.


-1

Interestingly, no exotic compound is required to consume the oxygen in a room, or, for example, a large sewer pipe. In fact, an interesting report of workers who died when drainage occurred in a sewer pipe, resulting in the exposure of fresh iron metal surfaces, likely in the presence of acidic conditions and a complexing agent. A reaction pursed resulting ...


1

One idea is to retry the brown ring test with the proper procedure, as the question admits that this had not previously been used (acidified ferrous sulfate was added wrongly). The other evidence is consistent with calcium nitrate, so we need to get that test right. Calcium nitrate also decomposes at 500°C and above, releasing nitrogen dioxide leaving the ...


1

The triethoxysilane group has to hydrolyze first in order to bond with a silanol group of glass surface. It seems that you are cleaning the slides very well. However, I think your temperature is too low and your time of exposure is too short. The hydrolysis of ethyoxysilane group is slow. Secondly, I guess you are posting questions on this topic for almost a ...


3

There are ways to quite reliably determine what this solid is. As mentioned above, the counter ion most probably should be $\ce{Ca^2+}$, however there are many organic or inorganic anions that still fit the criteria. The $\ce{AgNO3}$ tests provide insight that its a non-coordinating counter ion. Ways to determine the anion - however this all takes ...


4

Calcium formate is a white-to-yellow or off-white crystalline powder, has a density of $\pu{2.02 g/ml}$, solubility of $\pu{\sim 17\%}$, decomposes at $\pu{300 ^\circ C}$ . It irritates eyes severely and has a stinging taste, therefore could have an "hard to describe odor" or effect on the nose. One mental conflict I have is that I would expect ...


11

Very good question. I believe you are talking about ordinary Bunsen burner flame which is considered a low temperature flame by atomic standards. Good temperatures for atomic emissions are on the order of the temperature of the surface of the Sun (around 10,000 K). So you are right, it is very "difficult" to visually distinguish the apparent flame ...


35

Sodium flame is yellow, but all the light is due to two lines in the yellow region. If you look a sodium flame through a purple glass (cobalt glass), the yellow color is absorbed, and you do not see the sodium flame any more. The flame looks dark and colorless through a cobalt glass. On the contrary iron flame looks yellow, but it is made of a huge number of ...


6

I do not think the question had sodium dithionite. This is a mistake. It must be sodium thiosulfate, because sodium dithionite is such a strong reducing agent that it will not solubilize these metals. The properties mentioned in the answer match with thiosulfate. Sodium dithionite is such a strong reducing agent that it will reduce copper (II) to copper. ...


2

Following a discussion that took place in the comments: If given answers are correct, as Poutnik pointed out, your equation should be : $$\require{cancel}\ce{A(g) + B(g)<=>2 C(g)}$$ - Mathew Mahindratne Could you explain how you arrived at that conclusion? - Ollie This answer solves this question for a general case: $$\ce{a A(g) + b B(g) <=> c ...


1

Trial and error, but most often the boiling point of the lower-boiling solvent. The absolutely correct answer would be to look at a phase diagram, determine whether you have a positive or negative azeotope (or non at all), find out where on the phase diagram you are and then derive the temperature. That would tell you approximately at what temperature your ...


0

A galvanic cell with the same electrodes ( copper ones in this case ) is usable as a galvanic cell only if cathodic and anodic electrolyte differs in the concentration of electroactive components. E.g. if we can use 2 different concentrations of $\ce{CuSO4}$ solution, separating them by a diaphragm or a salt bridge. The electrode with more concentrated ...


2

By "gaseous titanium", do you mean "titanium tetraiodide ($\ce{TiI4}$) vapor"? Van-Arkel De-Boer process is a purification process of titanium and zirconium. Basically, what happens is that the impure metal, let's say titanium is heated in iodine environment at a temperature of $\ce{250 ^\circ C}$ to form volatile titanium tetraiodide ($\...


3

In order for a combustible substance to ignite, it must first attain a certain temperature called the 'autoignition point', which is the minimum temperature that the substance requires to begin combustion. In this case, the banknote didn't 'burn' because the water present absorbs the majority of the energy given off by the combustion of ethanol, which keeps ...


3

$\ce{NH3}$ and $\ce{SCl2}$ form $\ce{NH4Cl}$ which evolves as white fumes. $\ce{NH4Cl}$, when passed through $\ce{SCl2}$ (cherry-red liquid), gives the appearance of dense-pink fumes $$\ce{\underset{(cherry red)}{6SCl2} + 16 NH3 -> S4N4 +2S + \underset{(white)}{12NH4Cl}}$$ That is why the fumes of $\ce{NH4Cl}$ turned pink


1

In the case of excess of $\ce{H2SO4}$, bases like $\ce{Na2O}$ or $\ce{NaOH}$ cannot be produced. The first mentioned option ( $\ce{Na2SO4 + H2O}$ ) is not possible, as there would be oxidation without reduction. Note that in excess of $\ce{H2SO4}$, $\ce{NaHSO4}$ is formed, not $\ce{Na2SO4}$ Diluted $\ce{H2SO4}$ would produce $\ce{H2S}$, concentrated one ...


1

Well, first of all, it is important to understand the nature of this reaction. The reaction here is a redox reaction where one species is oxidised and another species is reduced. In this case, $S^{2-}$ is oxidised to $\ce{SO2}$ and $\ce{SO4^{2-}}$ is reduced to $\ce{SO2}$. Hence, even though $\ce{Na2SO4}$ does form (and it usually does since the Sulfide ...


16

When looking at IR spectra of hydrogen-bonding groups, always check how IR spectra was recorded. It appears that NIST spectra is recorded using gas phase, while the first one used liquid film. In liquid film $\ce{NH2}$ groups form a network of hydrogen bonds, resulting in shape distortion and widening of $\ce{NH2}$-related bands. Same is true for carboxylic ...


6

Honestly I have doubt about your textbook version of IR spectrum. However, I can suggest by the experience I have gained that the shapes of the peaks depends on the method you have used to obtain the spectra such as ATR, smear, absorbed in inert surface, etc., thus cannot pinpoint which one is incorrect. Yet, you can compare an one with other related amines. ...


2

Because of toxicity, I'd avoid middle or high school student experimentation with $\ce{CO}$. You might consider $\ce{H2O}$ and $\ce{H2O2}$, if you stick to 6% (20 vol) or weaker. There is the complication of working with a dilute solution, though, in getting precise measurements on proportions. Another possibility is demonstrating the various oxides of iron. ...


0

The freezing point remains the same, but its measured value may depend on the way of measurement. In case the solution freezing point is too close to the ice/salt 3:1 bath temperatures, it may be the reason for too long freezing time. The bath ice/$\ce{CaCl2 . 6 H2O}$ bath 1:2 could help, if applicable. Note that fast freezing may lead to biased value, ...


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