William H. Seward
(1801-1872)

American lawyer and politician (1801-1872) – William H. Seward was born in Florida (Village in Orange County, New York, United States) on May 16th, 1801 and died in Auburn (City in Cayuga County, New York) on October 10th, 1872 at the age of 71. Today William H. Seward would be 222 years old.

Age

how old was William H. Seward when he died?

71

William H. Seward died in 1872 at the age of 71.


Biographical data

Birthday May 16th, 1801 (Saturday)
Place of Birth Florida
Village in Orange County, New York, United States
Death Date October 10th, 1872 (Thursday)
Death place Auburn
City in Cayuga County, New York
Birth sign (Zodiac) Taurus (The Bull) ♉
Chinese Zodiac Rooster 雞

Other personalities born on May 16

Also born in 1801

William H. Seward

Birthday

When was William H. Seward born?

William H. Seward was born on May 16th, 1801.

Death

When did William H. Seward die?

William H. Seward died on October 10th, 1872 at the age of 71 in Auburn (City in Cayuga County, New York). Today William H. Seward would be 222 years old.

Place of Birth

Where was William H. Seward born?

Florida (Village in Orange County, New York, United States).

Death place

Where did William H. Seward die?

Auburn (City in Cayuga County, New York).

Zodiac

What is the zodiac sign of William H. Seward?

William H. Seward was born in the zodiac sign Taurus (The Bull).

Chinese Zodiac

Which Chinese zodiac sign does William H. Seward have?

William H. Seward was born in 1801 in the year of the Chinese zodiac Rooster.

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Died on October 10th, 1872:
Anniversaries of death

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American lawyer and politician (1801-1872)