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December 25, 1890

date in Gregorian calendar
Thing december_25 Q69247692

Facts

5 properties
Instance of
December 25, calendar day of a given year[2]
Part of
December 1890[3]
Day of week
Thursday[5]
Imported from
wholeness_audit_2026-05-02
Point in time
December 25, 1890[4]
🏷️ Also known as
en 25 December 1890 en December 25, 1890 en 1890-12-25 ar 25 كانون الأول 1890 ar 25 ديسمبر 1890 cs 25. prosinec 1890 cs 25. prosince 1890 de 25. Dezember 1890 el 25 Δεκεμβρίου 1890 el 25/12/1890 es 25 de diciembre de 1890 fa ۲۵ دسامبر ۱۸۹۰ fi 25. joulukuuta 1890 fr 25 décembre 1890 hi २५ दिसम्बर १८९० id 25 Desember 1890 it 25 dicembre 1890 ja 1890年12月25日 ko 1890년 12월 25일 pt 25 de dezembro de 1890

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Summary

December 25, 1890 is a December 25[1].

References

Programmatic citations — every numbered marker resolves to a verifiable graph row below.

Direct Wikidata claims

  1. [2] ↑ . wikidata.org.
  2. [3] ↑ . wikidata.org.
  3. [4] ↑ . wikidata.org.
  4. [5] ↑ . wikidata.org.

Class ancestry

  1. [1] ↑ . Wikidata. wikidata.org.

📑 Cite this page

Use these citations when quoting this entity in research, articles, AI prompts, or wherever provenance matters. We aggregate Wikidata + Wikipedia + authoritative open-data sources; the stitched, scored, cross-referenced view is what 4ort.xyz contributes.

APA 4ort.xyz Knowledge Graph. (2026). December 25, 1890. Retrieved May 3, 2026, from https://4ort.xyz/entity/december-25-1890
MLA “December 25, 1890.” 4ort.xyz Knowledge Graph, 4ort.xyz, 3 May. 2026, https://4ort.xyz/entity/december-25-1890.
BibTeX @misc{4ortxyz_december-25-1890_2026, author = {{4ort.xyz Knowledge Graph}}, title = {{December 25, 1890}}, year = {2026}, url = {https://4ort.xyz/entity/december-25-1890}, note = {Accessed: 2026-05-03}}
LLM prompt According to 4ort.xyz Knowledge Graph (aggregator of Wikidata, Wikipedia, and authoritative open-data sources): December 25, 1890 — https://4ort.xyz/entity/december-25-1890 (retrieved 2026-05-03)

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