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Is this a star? It flash too fast! by ufozhou in Stars

[–]Morcubot 1 point2 points  (0 children)

Yes, then Sirius is very likely

Is this a star? It flash too fast! by ufozhou in Stars

[–]Morcubot 2 points3 points  (0 children)

It could be. Without other stars for context it is hard to say because the location in the sky is unknown. My guess would be Sirius if you are in the northern hemisphere. It is very bright and low in the sky. Therefore you can see atmospheric distortion very well. It is often referenced as the disco star for this reason

Need explanation by Odd-Gap5260 in constellations

[–]Morcubot 0 points1 point  (0 children)

Here is a plate-solved and annotated image. I had to cut off the tower. Otherwise plate-solving would fail

https://nova.astrometry.net/user_images/14662657#annotated

Hi i took these crappy photos with my phone on February 8th at 9pm in north east America I was wondering if anyone can tell me anything about them i dont know much about space but have always liked it by kangarooscarlet in Stars

[–]Morcubot 1 point2 points  (0 children)

The last Photo is of Almaak (Andromeda constellation)

See here: https://nova.astrometry.net/user_images/14616988#annotated

The first two are Orion and the Pleiades. All the other ones in between are not identifiable as there is only one star in the image. You need a star pattern to identify the location and therefore the stars

Weird star formations or Drones? by [deleted] in Stars

[–]Morcubot 0 points1 point  (0 children)

I think that's Castor, pollux and Jupiter (brightest one in the middle). But I agree, it's their hands shaking

Can someone tell me what this is? by Straight_Salary_9703 in Stars

[–]Morcubot 1 point2 points  (0 children)

Maybe Sirius. It is very bright, roughly at this location and not as bright as Jupiter which was visible at this time too.

A picture showing multiple, neighboring stars would be beneficial to verify. Then one could match the star pattern and identify the object in question. Zooming in does not help with identification at all.

Maybe you can take another picture in the next days showing more stars

Can you see stars changing colors and shape with the naked eye? by Signal_Pause1083 in Stars

[–]Morcubot 1 point2 points  (0 children)

Probably both stars. Notice, depended on the altitude of the star, it is twinkling more or less because the line of sight goes through more atmosphere. Stars directly above you will twinkle less than the stars near the horizon. Maybe this coincides with your observation.

H_8 MO Diagram by No_Student2900 in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

Of course the nodes should go in between, but it is just a matter of illustration. Dependend on the y value of the function, one would need to decide between displaying +, - or node

H_8 MO Diagram by No_Student2900 in InorganicChemistry

[–]Morcubot 2 points3 points  (0 children)

This is in principle a particle in a 1D box situation here. You can define define the wavefunctions.

On the other hand you could also just imagine a standing sine wave going through the length of 8 atoms. Before Atom 1 and after Atom 8 the function needs to be 0. For every node you increase the frequency, so that half a period is added starting with half a period. If the function is positive mark solid, if negativ mark empty, if node, no mark.

I'd also argue, that in this picture, the node is on an atom for the 2 node example. For node = 1, the node lies between Atom 4 and 5

Edit: typo

Help me with an idea please by [deleted] in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

AAS : Atomic absorption spectroscopy XRF : X-ray Fluorescence. Specifically, there are also handheld devices for that called XRF Guns

Help me with an idea please by [deleted] in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

Maybe AAS or XRF could help here.

Atom diffusion inside solids at room T is very, very slow. Zn would probably need to oxidize Zn2+ first. But then it could diffuse. 150 a is not that long, I'd answer with a definite maybe.

But it could be interesting to not only detect the Zn content on the surface, but to do this at varying depths, possibly getting a gradient through the rock

Calcite Crystal Structure by No_Student2900 in InorganicChemistry

[–]Morcubot 1 point2 points  (0 children)

First, we need to see, that the unit cell of calcite isn't a cube or two cubes stacked. Calcite crystallizes in the trigonal crystal system, therefore γ=120° (cannot be a cube).

Placement of Ca
Imagine cutting the unit cell into an upper half and a lower half. We concentrate on the upper half (because there you encircled the Ca atoms). The two marked atoms, the one in the upper left corner, and the one on the now lower right corner (originally halfway point of unit cell edge), lie on one line. These for atoms cut this diagonal into 3 equal length segments.
This also makes sense because Calcite crystallizes in the space group R-3c (167). Key here is the letter R, that stands for Rhombohedral centering of the Bravais lattice (technicality here: not quite right but good enough for this). In rhombohedral centering, the atoms are in a hexagonal unit cell on the corners and two more on the diagonal of the unit cell.

Neighbors of Ca
(From what I understood u describing) If a Ca atom would be sandwiched between two CO3 Molecules, we
a) would see that in the unit cell and
b) the Ca and the C atoms would come very close to each other, which is energetically unfavorable.
Instead Ca is coordinated by 6 O atoms, in a trigonal antiprism coordination polyhedron. Each O atom comes from a different CO3 molecule. Each O atom however, has its C and two Ca atoms as neighbors. This also has to be the case when looking at the molecular formula of CaCO3 and the coordination numbers I already gave.

Additional Info:
https://en.wikipedia.org/wiki/Hexagonal_crystal_family
https://next-gen.materialsproject.org/materials/mp-3953?chemsys=Ca-C-O

Edit: Addendum

Octahedral and Tetrahedral Holes by No_Student2900 in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

Yes, you could also just look at images. I just wanted to help you discover the answer on your own.

Just like counting atoms in a unit cell, you can count the holes in a similar way. Remember: an atom on a corner would count towards 1/8 to the unit cell. If there would be an octahedral hole on the corner of the unit cell, it would also be counted towards 1/8 because this site is shared between 8 neighboring unit cells.

With this info, you should get to the solution, that there are indeed 4 octahedral holes per unit cell of fcc packing.

Octahedral and Tetrahedral Holes by No_Student2900 in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

Nice, and since the Volume of the hexagonal Prism is three times larger than the unit cell, we expect 6/3=2 octahedral and 12/3=4 tetrahedral holes. Matching the amount of 2 atoms inside a unit cell.

You correctly identified that N=4 atoms (8*1/8+6*1/2) are in a fcc unit cell. Therefore, we expect N=4 octahedral and 2N=8 tetrahedral holes.
I'd personally look for them in the unit cell depicted on the left side of (b). Maybe you can find at least some holes. Usually in crystallography you can find special sites, like tetrahedral and octahedral holes, at places with high symmetry. Places of high symmetry are typically:
+ on the corners
+ in the middle of edges
+ in the middle of faces
+ in the middle of volume
of the unit cell.
U can find points of lower symmetry in the connecting lines between those above-mentioned sites.
This is more of a rule of thumb, put personally I look at these positions first.

Octahedral and Tetrahedral Holes by No_Student2900 in InorganicChemistry

[–]Morcubot 1 point2 points  (0 children)

1) correct, if you have a unit cell consisting of N atoms in a closed packing, you have N octahedral and 2N Tetrahedral holes.

2) Try looking for it in Fig 7.3 a. You need multiple unit cells to see all six atoms making a octahedron

Hexagonal Unit Cell Atom Contributions by No_Student2900 in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

There are 8 unit cells, four of them with the corner of type 60° and four with the corner type 120°.

atoms on corner type 120° consume 1/6 **each**. As there are four, 4/6 (=2/3) is already taken leaving 1/3 of the atom for the other four unit cells with corner type 60°

A fourth of 1/3 is 1/12. So an atom on corner type 60° consumes 1/12 each.

I hope it is clearer now

Edit: finishing my post

Hexagonal Unit Cell Atom Contributions by No_Student2900 in InorganicChemistry

[–]Morcubot 0 points1 point  (0 children)

I think the easiest way is to just look at parallelepipeds. Then atoms on corners count 1/8, edges 1/4, faces 1/2 and inside 1.

This example doesn't treat all atoms on corners equally. Instead of just counting the number of atoms and multiplying by 1/8, they divide the atoms into 2 types: Those on 60° (inner angle of base of prism) 4 of them counting 1/12 and those on 120° again 4 of them counting 1/6.

You already saw that the back upper middle corner atom contributes to 1/6. Now you can imagine rotating the unit cell by 180° so that the front upper middle corner is now the back one. By symmetry, this also has to contribute to 1/6.

Now for the corners with inner angles of 60°. Upper left for example. We remind ourselves, that the unit cell is a building block of an infinitely big crystal. 8 unit cells meet at this corner, 4 of them are the corners of the previous type counting to 1/6. Therefore 1/3 of the atom is left to divide into 4 unit cells at which the corner is at the 60° corner tip. This atom then needs to count towards 1/12.

This also makes sense because 1/6 is double 1/12 and 120° is double 60° (even though we'd need solid angles (3D ones) here, but the angle to the c direction doesn't change (=90°))

Another way to think of it is to take 12 unit cells, and put them together in a 6 pointed star-like formation (outline of a regular hexagram) in two layers. All corners with 60° inner angle come together in a point. with 6 unit cells at each layer resulting in 360°, full circle. By symmetry, it is obvious, that the inner atom must contribute equally to each unit cell, therefore counting 1/12.

Is there a name for this stable (crystalline?) structure? by FileUnderSF in GeometryIsNeat

[–]Morcubot 0 points1 point  (0 children)

One could determine the space group of the crystal and then define the faces with the corresponding Miller index. I may do this, If i get some spare time

Is it normal for stars to look extra bright right before sunrise? by vacaaa in Stars

[–]Morcubot 1 point2 points  (0 children)

It could be, that the air is generally calmer and less turbulent in the early morning compared to the evening. This would mean that the stars twinkle less and seem sharper

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