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Jorge Escorihuela

@joresfu.bsky.social

Associate Professor (University of Valencia 🇪🇸) Computational chemistry and photocatalysis

36 Followers  |  122 Following  |  4 Posts  |  Joined: 07.10.2025  |  1.8224

Latest posts by joresfu.bsky.social on Bluesky

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La #UV presenta ‘29 O, Un any després’, el portal web que reuneix un any d'investigació, solidaritat i memòria després de la dana

🔗 Descobreix el portal: www.uv.es/29o

#29OUnAnyDesprés

28.10.2025 12:27 — 👍 3    🔁 2    💬 0    📌 0
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Rethinking peer review in the AI era with responsibility and transparency In a rapidly evolving AI-driven landscape, we explore how AI can support, but why human judgment must remain in the driving seat

Rethinking peer review in the AI era with responsibility and transparency
www.elsevier.com/connect/reth...

28.10.2025 14:17 — 👍 0    🔁 0    💬 0    📌 0
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Fallecimiento del Prof. Miguel Ángel Miranda Alonso – RSEQ Lamentamos comunicar el fallecimiento del Prof. Miguel Ángel Miranda, catedrático jubilado de la UPV e investigador del ITQ.

La @rseq-quimica.bsky.social expresa sus condolencias a la familia, amigos y colaboradores del Prof. Miguel Ángel Miranda Alonso, catedrático recientemente jubilado de la Universitat Politècnica de València (UPV) e investigador del Instituto de Tecnología Química (ITQ) rseq.org/fallecimient...

24.10.2025 18:07 — 👍 3    🔁 3    💬 0    📌 0
Infographic on the mole. One mole is the amount of substance that contains exactly 6.022 x 10^23 atoms, molecules or ions. This number is also known as Avogadro's number. Using moles makes it easier to talk about amounts of substances involved in reactions by relating the mass of a substance to its atomic or molecular mass. Amount of substance (moles) = mass (grams) divided by the mass of 1 mole (grams per mole). One mole contains a different mass for different substances.

Infographic on the mole. One mole is the amount of substance that contains exactly 6.022 x 10^23 atoms, molecules or ions. This number is also known as Avogadro's number. Using moles makes it easier to talk about amounts of substances involved in reactions by relating the mass of a substance to its atomic or molecular mass. Amount of substance (moles) = mass (grams) divided by the mass of 1 mole (grams per mole). One mole contains a different mass for different substances.

I'm a bit late with the customary #MoleDay infographic this year, but better late than never!

The mole makes it easier to describe the huge numbers of atoms, molecules, or ions that participate in chemical reactions.

More in this graphic: www.compoundchem.com/2014/10/23/m...

#ChemSky 🧪

23.10.2025 19:34 — 👍 113    🔁 38    💬 5    📌 5
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Predicting the Optical Properties of Gold Nanoclusters Using Machine Learning Approach The synthesis of gold nanoclusters (AuNC) is strongly influenced by various reaction conditions, and their optical properties are determined by factors such as the nature of the ligand and the measuri...

Check our new publication in ACS Omega | Predicting the Optical Properties of Gold Nanoclusters Using Machine Learning Approach | pubs.acs.org/doi/10.1021/...

21.10.2025 19:00 — 👍 2    🔁 0    💬 0    📌 0
ball-and-stick and line structures of a porphyrin nanobelt

ball-and-stick and line structures of a porphyrin nanobelt

Lots of great chemsky content in @science.org this week, including a remarkable porphyrin nanobelt from the Saywell and Anderson groups 🧪

www.science.org/doi/10.1126/...

16.10.2025 18:14 — 👍 81    🔁 13    💬 4    📌 7
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The 2025 Nobel prize in chemistry as it happens – live Join us as we provide analysis and commentary in the run up to the announcement of the biggest prize in chemistry

The 2025 Nobel prize in chemistry as it happens – live
www.chemistryworld.com/news/the-202...

08.10.2025 10:17 — 👍 0    🔁 0    💬 0    📌 0
https://www.nature.com/articles/s41467-025-61819-6

A beautiful paper in Nature Comunnications from our collaboration with Prof. Norio Shibata from Nagoya Institute of Technology t.co/43BeMYWE0B @icmoluv.bsky.social @uv.es @nitechno.bsky.social

08.10.2025 09:02 — 👍 1    🔁 1    💬 0    📌 0

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