New answers tagged inorganic-chemistry
1
Tl, dr: Upon further review the differences between beryllium, magnesium and other metals is not so clear-cut.
There is less to the difference in flame colors than meets the eye. We do not have polar opposites of colored flames at one end and white flames at the other. Potassium does not introduce a strong color to the flame even though we often call the ...
4
For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find ...
1
The other answers provide valuable insight into what actually happens and are very helpful in understanding the chemistry behind it.
But to correctly answer the question, a much more basic understanding of chemistry (and how tests work) would suffice.
Let's look at what is given in the Question:
Nickel sulfate, NiSO4(aq) is a green solution. Nickel ...
2
Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given)
If no information is given, you should just give the three amounts names and treat them as unknowns.
ZnO is exposed to pure CO at 1300 K
This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental ...
1
Tetrakis(triphenylphosphano)palladium(0) or $\ce{[Pd(PPh3)4]}$ is a common catalyst in organic chemistry that features a palladium(0) core with four triphenylphosphane ligands. It is a bright yellow, sometimes described as canary yellow. See the image below (taken from Wikipedia, where a full list of authors is available); however, I feel the picture doesn’t ...
3
Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution,
$$n_1 = n_2 $$
Since $n = MV$,
$$M_1 V_1 = M_2 V_2$$
The second is related to the conservation of the total number of moles in an isothermal compression or expansion,
$$n_1 = n_2$$
Using the ideal gas law $n = pV/RT$,
$$\frac{P_1 V_1}{...
8
Jan already gave a great answer explaining the real chemistry behind this, which you and your daughter should absolutely read. This answer also makes the point that the question is poorly worded and could be contested on that basis—that is absolutely true, but unfortunately rather common in high school chemistry courses.
In high school chemistry, one ...
2
It reacted with the baking soda and is no longer sulfuric acid, but much safer sodium sulfate. If you're unsure, add more baking soda until it no longer fizzes.
The remaining white powder is a mix of baking soda or $\ce{NaHCO3}$, and sodium sulfate, $\ce{Na2SO4}$. It safe to wipe up with a damp rag and fairly harmless to touch. Just avoid getting the ...
17
The question is really badly worded. For starters, let’s look at solutions of nickel(II):
Figure 1: Nickel(II) solutions. From left to right: $\ce{[Ni(NH3)6]^2+}$, $\ce{[Ni(en)3]^2+}$, $\ce{[NiCl4]^2-}$, $\ce{[Ni(H2O)6]^2+}$. Image taken from Wikipedia, where a full list of authors is available.
You can ignore the left two but the rightmost is a standard ...
2
$\ce{HSO4-}$ is acidic ($\mathrm pK_\mathrm a = 1.99$) yet also reacts as a nucleophile e.g. in the acid-catalysed hydration of olefins, which according to my mechanism lecture procedes via a sulphate diester intermediate.
If I thought for long enough, I probably could come up with a basic compound that acts as an electrophile in a given situation.
However,...
7
Anhydrous nickel chloride is yellow. However most simple divalent salts of nickel are green. This should be good enough for high school. The question is poorly worded because it is ambiguous (hopefully the textbook will improve it). If they discuss metal complexes it can be mentioned that $\ce{[Ni(H2O)6]^2+}$ is green in color.
3
Most metal nitrates decompose on heating to give metal oxides. Decomposition of silver nitrate is different as it yields elemental silver.
$$\ce{2AgNO3 -> 2Ag + O2 + 2NO2}$$
Usually the metals which lie very low in the activity series decompose to give their elemental form.
I don't think it's possible to extract metals from just any nitrate by strong ...
0
Cryollite is used to decrease the temperature from $2000 °C$ to $900 °C$ . Fluorspar is used to increase mobility by controlling viscosity, hence in turn increasing conductivity.
1
$$\ce{Mn^{2+} + Cr2O7^{2-} -> Mn^{3+} + Cr^{3+}}$$
If you were to balance this reaction, using the appropriate molecules of water and protons, you'd get:
$$\ce{ Mn^{2+} + Cr2O7^{2-} + 14H+ + 5e- -> Mn^{3+} + 2Cr^{3+} + 7H2O }$$
The presence of protons the left side of equation suggests that this reaction has a higher rate when the concentration of ...
4
All metals look black or dark when presented as a sufficiently thin powder. The metals display their color only when the dimension of the grain is greater than the wavelength of the light, namely 400 to 700 nm. This is about 1000 times bigger than the atoms. When prepared from a precipitation reaction, the metal atoms are never arranged neatly in a regular ...
8
While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at ...
2
Zinc will amalgamate with mercury, but whether that reduces the vapor pressure of mercury very much is debatable. At best, the increased volume of zinc amalgam and the ability to amalgamate with fine droplets of mercury would make it easier to clean up the mercury as amalgam.
It might be better to spread some sulfur dust (flowers of sulfur) over the area. ...
1
To add to Alchimista's answer, suppose the question were about a sample containing 1 mole of ozone (and nothing else). You could ask two questions:
What is the mass of ozone in the sample?
What is the mass of oxygen in the sample?
Because all the oxygen atoms are in the form of ozone, the answer to questions 1. and 2. would be the same. On the other hand, ...
3
The question in the book is about elemental composition of the given substance, in terms of mass.
You are given the empirical formula, i.e. what could be obtained by elemental analysis, namely what elements composed the substance and in which relative ratio. This does not tell you about the spatial arrangement of the various elements and the way they bound ...
3
Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied.
However, there is no requirement that a path between two states ...
1
$\ce{CuSO4}$ exists as $\ce{[Cu(H2O)4]SO4}$ in solution. It is blue in colour due to the presence of $\ce{[Cu(H2O)4]^{2+}}$ ions.
Now if aqueous $\ce{KF}$ is added, the solution turns green due to formation of complexes $\ce{[CuF4]^2-}$ and $\ce{[CuF6]^4-}$(Initially copper(II) fluoride($\ce{CuF2}$) is formed but it formed complexe in present of water). The ...
2
A battery has a small shelf life if it discharges internally. Saying that a discharged battery is in chemical equillibrium (or much closer to it than a fresh one) is a true statement, obviously, but nothing else.
If you reverse the polarity of a primary battery, you get some other reaction than the one happening during discharging backwards. That says ...
1
If the calcium chloride is already anhydrous (as shown by the amount of water, $\ce{H2O}$, by weight percent), i.e. has little water, then there's nothing to do but package it in a manner that when it liquefies from absorbing water from the air, it won't drip on containers below. For an example, with a plastic box or pail below.
If there is substantial ...
1
Both equations are balanced, both equations are correct. Using other coefficients such as $0.5, \sqrt{2}$ or ${e^3}$ would also be correct. Only if the ratio of coefficients were different (i.e. not $1:1:1$) it would be wrong.
However, it is commonly preferable to write the smallest possible integers, so $1,1,1$ is preferred over $2,2,2$.
2
The thermal decomposition of siderite is interesting aspect. First, the answer to your question is your textbook has mistakenly put a common factor (of 2) to balance the chemical equation. It is not wrong but traditionally we do not put a common factor on equations, unless we intended to add or subtract two of more equations. Said that, I have a doubt that ...
4
Well, just like any other reactions, it depends upon the the heating temperature and the reaction environment. Contrary to what the reaction tells, the decomposition products of the reaction is actually iron(II,III) oxide, $\ce{Fe2O3}$ if it is heated in oxygen environment$\ce{^{[1]}}$ and $\ce{Fe3O4}$ if it is heated in $\ce{CO2}$ environment$\ce{^{[2]}}$. ...
2
Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.
Step 1: Start with the given metal species, with the metal oxidation states included:
$\ce{M^VO4^{3-}->M^{III}O(OH)}$
Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there ...
6
This is a misprint. Here is a similar problem (OpenStax Chemistry, retrieved from https://opentextbc.ca/chemistry/chapter/7-4-formal-charges-and-resonance/) that makes sense and has a correct answer:
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular ...
0
After searching on the internet, and going through textbooks, I understand there is no "shortcut-method" to get the atomic term symbols in atoms with a large number of electrons.
4
I always suggest students to try Google Scholar (scholar.google.com) when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant.
When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google ...
1
The explanation given to you is rather incorrect. Dilute sulfuric acid is added to iron(II) or even iron(III) salts is added to prevent hydrolysis of the salts. What happens when you don't add an acid to iron(II) solution in water. With time, you will see that a brown precipitate is forming and settling to the bottom. Iron forms hydroxides which are very ...
1
Resonance structures can only be drawn for single compounds. Rephrased: if a set of resonance structures are to be valid, then all atoms (nuclei) must be at exactly the same positions.
Thus, when asked to draw resonance structures of e.g. phenol or nitric acid, you cannot deprotonate or transfer protons. All protons must be exactly at the same spot in all ...
2
If you have seen pink salt crystals from Khewra mines in Pakistan (the so-called Himalayan pink salt) sold at exorbitant rates in grocery stores, you would find out it is nothing but NaCl with a small amount of iron oxides as stated above. If you dissolve large tablespoons of such salt crystals in water, you will see a brownish precipitate sitting at the ...
5
It depends on what specifically the question is asking you to do.
In reality, a polyprotic acid such as $\ce{H3PO4}$ becomes deprotonated in a sequential way; the first proton is the easiest to remove (most acidic), then the second, then finally the third is hardest to remove (least acidic).
So there are four different species that the question might be ...
5
This question serves as a prime example of how much fake and scientifically incorrect information, or more famously, fake news is there on the web. This is why ordinary web should not be anyone's first priority to search about scientific topics. There is a famous saying that "paper never refused an ink" which applies to the web as well in the sense that ...
2
A previous comment noted that sodium compounds are generally soluble and would not be strongly adsorbed on malachite and anglesite.
However, the other compound formed in the reaction between Na2S and a copper or lead salt would be CuS or PbS, both of which are very insoluble. Three cases can be visualized: 1) a sodium ion approaches the Pb or Cu salt and ...
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