# Tag Info

In multielectronic atoms we have a relatively large difference between $s$ and $d$ orbitals when they have the same $n$ quantum number, or in terms of the actual quantum mechanics when they have the same total number of nodes in the wavefunction (this being what we label as $n-1$). But in the case of transition metals the $s$ orbital that mixes in with the $... 1 A number itself doesn't inherently precision or accuracy. It is the representation of a measurement: the output of a measuring instrument. That instrument is what has precision and accuracy. We accept the number representing an instrument's output on the basis that we trust the calibration and quality of the instrument. Establishing a certain number of ... 4 While estimating uncertainty can be quite involved, the purpose of significant figures is quite simple: They are a shorthand method of communicating the uncertainty of a measurement from those who are in a position to be able to estimate the uncertainty (those who made the measurement) to those who need to know the uncertainty, but can't determine it ... 2 Run the following demos with a cup of water in the microwave, and the test object on an inexpensive glass or ceramic plate, to reduce the risk of damage to the microwave itself. Even with care, expect plates to crack, and there is still possibility of damage to the oven. You might also want to remove or cover the light built into the microwave oven to see ... 5 Once upon a time, calculations were done on paper and on slide rules. It would take considerable effort to carry 10 digits in a calculation instead of 3 (which is what most slide rules are capable of). Then, students started using calculators, and would just copy the result they got, with 10 or more digits. Science teachers thought that was ridiculous, and ... 5 You are absolutely correct in that there are several kind of uncertainties (often called errors). These are referred to as systematic and statistical errors and the difference between them can be conveniently be described in terms of a target with a distribution of shots. The distribution of the shots with respect to the bullseye represents the kind of ... 1 Assuming you want to generate vapors of iodine at ambient pressure, you need to surpass its boiling temperature. Wikipedia's box about iodine's physical properties state about$\pu{184 °C}$for this phase transition ​(about$\pu{364 °F}$). You intend to use a barrel made of high density polyethylene, HDPE. Wikipedia's entry for a typical specimen of HDPE ... 3 The alpha particles used in this experiment have an energy of about$5.59\ \mathrm{MeV}$. This energy is too high; the alpha particle cannot simply capture two electrons and form a stable helium atom. By way of comparison, the first and second ionization energy of helium is$24.6\ \mathrm{eV}$and$54.4\ \mathrm{eV}$, respectively. Nevertheless, alpha ... -1 An alpha particle is basically a helium nucleus, with two positively charged protons and no electrons, the particle would react with any electrons it comes into contact with. However it likely depends on which electron it interacts with, and how close the particle comes that determine the outcome. As ScienceDirect describes here - In some cases, this force ... 2 The short answer is no, as such this task is not possible. You mention$\ce{Pd2(dba)3}$and$\ce{P(o-tol)3}$dissolve in the solvent of reaction; this describes an example of homogeneous catalysis, where the reagents share a phase in common. A simple filtration to remove the Pd catalyst will not work. It is like passing water of the ocean through a paper ... 1 If possible, you may use Whatman filter paper. It has specially been designed for this purpose. Well, that kind of a paper is preferable which has a good pore size, kind of a paper of unglossed newspaper. 2 The boiling temperature always increases when adding salt into water. It looks as if salt was attracting water molecules and preventing them from quitting the liquid. So that it is necessary to overheat the solution to let water molecules get into the vapor phase. I have often made this measurements in class and always found about$\pu{104°C}$. 2 The question lacks to specify e.g, the type of X-ray radiation (energy of exciting electrons), or the nature of the target. In parts, arguments put forward by @Amir Kalashnikov may be misunderstood as if concepts of classical optics (e.g., for a reflection target, as if the angle between the incident ray and the surface normalwould equate the angle ... 2 Soap is a fatty acid salt of an alkali metal, typically sodium or potassium. The fatty acid itself is always insoluble in water, and therefore acidification of the soap solution produces two layers, a lower aqueous layer with the alkali salt of the acid and an upper layer of fatty acid laced with aqueous material. This can be treated with an organic solvent ... 2 The angle of resulting X-ray radiation, product of the impact, is related to the amount of energy or momentum of the cathode ray. Bremsstrahlung relation will amount how much energy the produced photon has. $$hf = \frac{hc}{\lambda} = K - K'\tag{1}$$ $$\lambda_\mathrm{min} = \frac{hc}{K} = \frac{hc}{e\,\Delta V}\tag{2}$$ Back scatter the possible photon ... 5 Usually, if people have experience of boiling water at home and think a while about it, they are able to answer it themselves. These "boiling tubes" are used primarily for ambient temperature testing, so they are called test tubes, or test-tubes. Taking a test-tube in and out is for decreasing the heat flow to prevent violent overheating and the ... 1 I am doing another approach, to get another result. Starting from$\pu{0,1158 g}\ce{Co3O4}$, whose molar mass is$\pu{241 g/mol}$, this amount corresponds to$\pu{\frac{0.1158 g}{241 g/ mol} = 0.4804 mol}\ce{Co3O4}$, and also to$\pu{3*0.4804 = 1.4415 mmol }\ce{Co}$. This amount of cobalt is included in the original sample of oxalate, which weighs$\...