18

The analytical technique is called inductively coupled plasma optical emission spectroscopy. It was invented by a chemist V. A. Fassel, although it is pure physics. The specimen is digested in an acid, and a fine spray is created. That "spray" is heated to a temperature, which is higher than the surface of the Sun. Most of the elements of the periodic table ...


11

There are many ways to detect mercury, and you can find a more detailed analysis of these methods, along with examples of these detectors and their limitations, here. Gold film sensors Gold Film Sensors were the first reliable forms of mercury detectors due to gold’s affinity for elemental mercury [...] When a mercury rich air sample passes over a thin ...


11

Most commonly they work by pulling air through a UV absorption cell and simply measuring the absorption at 254 nm. Often there is a pre-concentration step involved. This usually involves a gold-plated region prior to the UV cell. At room temperature mercury adsorbs to the gold, then after some period of time the gold-plated region is flash heated to ...


11

There is no chemical symbol for paper since it is not an element but rather a mixture of several different compounds. Remember that paper is made out of trees. Thus, paper is made mostly out of organic compounds: that is carbon, hydrogen and oxygen (C, H and O). Paper also contains non organic materials to improve its properties. These may be chalk (CaCO3) ...


11

Inductively Coupled Plasma (ICP) based techniques are very common for elemental analysis. The goal of such analyses is to quantify the elements present in a sample. The neat thing about this, in my opinion, is that the integrity of the molecular compounds does not need to be maintained throughout the analysis, since we only care about the elements that ...


9

TL;DR f-f transitions in lanthanide ions give rise to the colours of the ions. The electronic states can be derived from quantum mechanics, and their energies vary from ion to ion, which give rise to different colours.. The colours are pale because these transitions are Laporte-forbidden and therefore weak. These absorption bands do indeed arise from ...


9

Aluminium(III) chloride - often described as white but samples is sometimes contaminated with iron(III) chloride giving it a light yellow color Antimony(V) oxide - pale yellow solid Bismuth(III) oxide - light yellow solid Rhodium(III) oxide - lemon yellow Cerium(IV) oxide - pale yellow-white compound Dysprosium(III) nitrate - white to pale yellow Holmium(III)...


7

$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$ Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$: $$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$ So the ...


7

Disclaimer: Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school. For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen. Employing a bit of basic math in the analysis of that ...


7

Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have $$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637 }$$ Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R ...


6

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types: Point defects Line defects. Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one. One type of point defect is a vacancy defect in which ...


6

Neutron activation analysis is a technique that can be used for trace mineral quantification in a sample. You start by irradiating your sample in a small experimental nuclear reactor. These were more common in the 1960's and '70's than they are today. Slow neutrons are absorbed by many different isotopes, converting them to radioactive species. The sample is ...


5

If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.


5

You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is: $$IHD = \frac{2C + 2 - H - X + N}{2}$$ Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + ...


4

First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate: cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$ hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$ lignin: $\ce{C31H34O11}$ with $M_l = 582~\...


4

Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated. From [1, p. 358]:...


4

When you ask how to remove borax, I assume it is Sodium tetraborate decahydrate (Wikipedia). Thus, I agree with @andselisk of using $\ce{Ba^2+}$ solution to precipitate it. Even though it is seemingly an excellent answer, the reference given and the text body are in German, and I didn't understand it much. :-) Thus, I want to give some clues to make it sense....


4

Calcium hypochlorite is an example of such a yellowish white compound. Another example would be Silver Bromide


4

The reaction of any acid on a thiosulfate solution makes a precipitate of sulfur $\ce{S_8}$ which is pale yellow, nearly white.


3

It will depend on the size of the particles which are in suspension. If the particles are very large compared with the capillary feeding the solution into the spray chamber then the capillary will block up and you will detect nothing and end up very unhappy with a broken machine. If the particles are much smaller than the capillary then they will be able to ...


3

You can check if the calculated molecular weight corresponds to lactic acid if not try next multiple like 2, then 3 etc. You get $\pu{30g/mol}$ for $\ce{C1H2O1}$ and $(\pu{90g/mol})/(\pu{30g/mol})=3$ So the multiple to be applied is 3: $\ce{C3H6O3}$


3

I suppose that you are referring to a Wiley–McLaren time‐of‐flight mass spectrometer [Rev. Sci. Instr. 65, 3344 (1994)] which is basically a tube containing an electrode stack for accelerating the ions toward a detector. As mentioned in the comments, one can discriminate different masses if one uses a pulsed experiment, for instance using pulsed lasers or ...


3

A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), ...


3

For the determination of calcium (at least) two methods are conceivable: gravimetric analysis Sulfate is a rather bad choice here. As compared to the carbonate, the sulfate has a much higher solubility. A better choice would be the oxalate with a solubility around $6\,\mathrm{mg\,l^{-1}}$. photometry Calcium forms a purple complex with the Arsenazo III dye ...


3

To best understand, let us consider the reactions occurring. First, we have $\ce{Cl-}$ ions in solution. Acidifying the solution serves to remove other anions that might give a false positive for the chloride test. Upon addition of $\ce{AgNO3}$, the following reaction occurs: $$\ce{Ag+(aq) +Cl-(aq) -> AgCl (s)}$$ When enough ammonia is added to ...


3

I did this a little differently. If there were $\pu{4.5897 \times 10^{-3}}$ moles of $\ce{CO2}$ produced, that must mean that there were $1/7$ as many moles of $\ce{C7H16O4}$ present in the original sample, or $$\frac{4.5897\times 10^{-3}}{7}\times 164.1995=\pu{0.10766 g}$$ in the original sample. The rest of the original sample ($\pu{0.01469 g}$) must ...


3

The comments more or less say it all. Wearing gloves is a good idea with just about any chemical, not only to protect you but also to protect the chemical from contamination. For instance, if you find chloride in your sample you might want to be sure said chloride is not from salt on your skin.


3

The test for Chloride ion is silver nitrite in an Acidified solution [with nitric acid]. This removes carbonate (as carbonic acid), cyanide[careful] and sulfide (as hydrogen sulfide). Sulfate must be absent, if present it can be removed with barium nitrate (as barium sulfate). The silver chloride formed is solubilized by addition of ammonia; insoluble ...


2

It depends what you mean by "exactly". A single gram of salt contains about 1022 atoms of sodium and chlorine. The important question is how much of an imbalance would be physically noticeable. If you can't detect the difference it isn't significant. A single atom discrepancy would be impossible to detect with the best equipment available to science. So it ...


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