Lots of confidently incorrect answers on here. When you specifically mention a competitor brand, you need to have legal proof of your claim, and that proof may need to be made publicly available to your competition which in turn may reveal trade secrets to them, e.g. the ingredient formula that allows you to make a claim that you have “better taste” or the brewing process that leads you to be “less filling.”

So it becomes a cost-benefit conversation. Is the proof behind your claim worth revealing to your competition?

Soylent Grain

But seldom by name. Always by his own nicknames for them.

Which is why the transition metals start in the fourth period, but utilize the 3d orbital

Always punch up never down

How old are you? If you’re still young keep an eye on that.

“Valence” refers to the highest level of orbital(s) where electrons relatively freely wander between structures with similar energy levels. How many electrons this is will vary with atomic size.

1s is the lowest orbital, and has no friends - so helium’s valence shell is just two electrons.

Then there’s a big jump.

2s is similar in energy to 2p, and so electrons will move between the two. To fill this next valence layer you need eight electrons.

Then there’s a big jump

3s is similar to 3p, again giving you an 8-valence level.

Then there’s a big jump.

4s is where you’re now at energy levels high enough to generate a d-orbital, and 4s, 3d, and 4p form all sorts of wacky structures. This layer has 18 in the valence level.

Then… you guessed it… there’s a big jump.

“Full valence” is where these big jumps occur. Adding or subtracting an electron from a full structure is relatively difficult. You’re either forming an entirely new structure on top of the existing one or destabilizing a fully spin-paired configuration. Both are difficult, and so atoms and ions with full valence are highly chemically stable.

> I know this, im just wondering if d is a part of the valence shell of that energy level.

The 3d shell is on the same level as the 3s and 3p orbitals in a hydrogen atom, but in the larger atoms that actually fill the 3d shell in their ground state, it turns out that the 3d shell ends up much higher-energy than the 3s and 3p shells because of how it interacts with other electrons. It ends up with an energy between the 4s and 4p orbitals instead, so for the purposes of the periodic table, you can think of 3d as hanging out in the fourth "shell" instead.

The notion of "shells" is a simplification.

You are familiar, I imagine, with the idea of the s-, p-, d-, and f-orbitals, since you mention them in your post. These correspond to ℓ = 0, 1, 2, and 3 respectively in the state of an electron. And they come in different levels, given by the value of n, also part of the description of an electron. So for example, the 2s subshell corresponds to n = 2 and ℓ = 0.

As a broad rule, these subshells are filled in the following order:

• Subshells with lower n + ℓ are filled first.
• For subshells with the same n + ℓ, start with the lowest n.

That results in the order 1s, 2s, 2p, 3s, 3p for the first three rows (with n+ℓ values of 1, 2, 3, 3, 4, 4 respectively). But once you get to the next row, the first that contains d electrons, it goes 4s, 3d, 4p. It doesn't "skip" d, it's just that the d it's filling in the fourth row of the periodic table is the 3d subshell, not the 4d one. The noble gas in that row (krypton, as it happens) does in fact have its highest occupied d shell filled. It has the electron configuration [Ar] 4s^(2) 3d^(10) 4p^(6) - this 3d subshell is full.

The reason the 3d shell shows up between 4s and 4p here, even though the n is nominally an energy level, is that that these numbers describe the energies of an orbital in the absence of other electrons. But other electrons in the atom jostle energy levels quite a bit. It turns out that when lower orbitals are occupied, d- orbitals end up so high energy that they effectively get "bumped up a tier" of the table.

It doesn't have any 4d electrons yet, because 4d electrons would be much higher energy than the 4s, 3d, and 4p electrons it actually has.

The better way to think about this is in terms of the gaps in energy levels. Noble gases have a large gap between the energy level of the highest-energy electron they have and the next available electron slot. That makes sticking an electron to them hard, because that electron has to occupy a high-energy state. And it makes stripping an electron off of them hard too, because all the electrons they have occupy low energy states. Their configurations look like this (where the blue lines represent energy of occupied orbitals, and red represents unoccupied orbitals).

As you go down the periodic table, the notion of "shells" starts to become less useful, because the gaps between the shells shrinks enough that the gaps within the shells can start to cause them to spill over one another. So chemistry near the bottom of the periodic table becomes more complicated, and in fact it's generally believed that if element 118 - which is one of the noble gases by its position on the periodic table - would actually be a solid if it were stable enough to stick around and have any chemistry at all.

I know this, im just wondering if d is a part of the valence shell of that energy level. People say that noble gases have full valence shells but us ut really full if d sublevel is skipped?

For the 3rd level the noble gas only full s and p levels, skipping d sublevel so the valence shell is not truely full for noble gases past argon right?

Woah

Yes

So to put it in ELI5 terms, someone can take a Bitcoin, wrap it, sell the WBTC on Ethereums chain and reverse the transaction to get back their BTC while profiting off of the WBTC sale?

> if you are the category leader, like say Coke over Pepsi -- you NEVER, ever mention your competitor.

One of my favourite ad campaigns ever is this from Pepsi v. Coke:

https://www.brandme.com.au/blog/2014/10/coke-vs-pepsi-a-scary-halloween-ad-campaign/

In the UK media the pound is usually described against both the dollar and the euro, as those are the two biggest currencies for UK trade, and it's not uncommon for them to go in different directions.

There is no 1p, 1d, or 2d structure. These atoms are too small and compact to support the more complex and diffuse electron configurations required to stuff dozens of them into the same space.

The 3d does exist, but it’s higher energy than the 3p and 4s and so doesn’t fill in until later.

Eventually you’ll fill an f-orbital too.

But isnt the dshell part of the valence shell of that energy level? I know you skip to the s sublevel of the next level before going back to d. Does this mean the level is not a part of the valence shell for that particular level?

The heavier noble gases do have a full d-shell as well.

You can fill four successive s orbitals before a d-shell configuration becomes energetically favorable, so there is no d-shell for Helium, Neon, and Argon.

Krypton has a filled d-orbital. Helium only has a full 1s, no p or d.

There are a few loopholes.

51% is an obvious one- send bitcoin to get WBTC and then reverse the transaction on the bitcoin chain. Sell the WBTC to double your "money". It's one of the big defi piñatas.

Not happening with bitcoin any time soon (we'd need several more halvings), but you can see it with smaller tokens.

Also the usual boring hacking stuff like input validation and social engineering.

> From my (limited) understanding, entangled particles don’t remain entangled if they are far apart or if something else “touches” one of them, which is not at all how it is explained in the article.

Entanglement is not an all-or-nothing deal. Entanglement is correlation + superposition. If two particles are in a 'large' superposition (they are simultaneously in multiple states which are very distinct from each other), and their states are highly correlated with each other, they are very entangled. This is what typical spin entangled states look like. One particle is 50% spin up and 50% spin down, the other is 50% up and down, and they are maximally correlated - they will always be spinning in opposite directions. When you observe one particle, you precisely know the state of the other.

However, it's also possible to have very weak entanglement. For example if one particle is in a location superposition, and it interacts with another particle from a distance, their trajectories are now entangled. But this entanglement can be arbitrarily weak. If you observe the location of one particle, you gain a tiny amount of information of the other one - if you thought the location would be clustered around some target, you know that cluster might have shifted a little bit. But it doesn't particularly help you narrow down the location.

There's no strict line between weak entanglement and no entanglement. Extremely weak entanglement still exists in macroscopic interactions. Our classical world arises from continuous weak entanglement of the environment correlating everything's state, so we don't see any disconnected parts of the world existing in large superpositions.

That's what the skeptics are saying. The tardigrade is entangled with the qubit in a very weak sense. If you had a godlike view that could measure the exact quantum state of every particle in the tardigrade's body, you would notice a very slight correlation between them and the state of the qubit. The tardigrades are not in a clean set of two possible states that exactly correspond to the qubit, they are in the same multitude of microstates that continually interact with surrounding objects and some probabilities of those microstates have shifted around just a little bit.

There have been experiments that do actually entangle macroscopic objects, e.g. large mechanical oscillators. The 'Results' section is the most informative - not only does there have to be interaction with some microscopic entangled states, but the states of the oscillators should also be highly correlated, and the statistical properties have to demonstrate that they are in fact in a non-classical superposition. The tardigrade experiment did not do this clear work.

Cannibal juice. And barley

Yes, I didn't mean that the idea in itself was bogus, but they had no way to actually validate that anything happened.

This high level description isn't fundamentally different to how you would actually entangle macroscopic objects. We know how to create microscopic superposition states, and we know that entanglement spreads by interaction. That's the point of the Schroedinger's cat experiment - the decaying atom interacts with the cat through the detector+poison vial, and this interaction entangles them, putting both in a very distinct superposition of vastly different states (decayed and not decayed, and dead and not dead).

The question here is whether the tardigrades were meaningfully entangled with the qubit states by this specific interaction (acting as a dielectric on capacitors within the quantum system). The skeptics say that the interaction of the tardigrades with the actual qubit states is so weak, there is effectively no correlation between the two.

Look at it's value relative to other key currencies like the Euro, Pound, Yen. If your dollar buys fewer Euros then the dollar is weakening, if you can buy more Euros then the dollar is strengthening.

There are quite literally markets where money is bought and traded, people literally just saying "I'll give you X of this currency for Y of that currency." But taking an average of these markets, you can determine the relative value of any currency to another at any given time.