Vice president of the United States from 1801 to 1805, lived (1756–1836) – Aaron Burr was born in Newark (City in Essex County, New Jersey, United States) on February 6th, 1756 and died in Port Richmond (Neighborhood on Staten Island) on September 14th, 1836 at the age of 80. Today Aaron Burr would be 268 years old.
Aaron Burr died in 1836 at the age of 80.
Birthday | February 6th, 1756 (Friday) |
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Place of Birth | Newark City in Essex County, New Jersey, United States |
Death Date | September 14th, 1836 (Wednesday) |
Death place | Port Richmond Neighborhood on Staten Island |
Birth sign (Zodiac) | Aquarius (The Water-Bearer) ♒ |
Chinese Zodiac | Rat 鼠 |
Aaron Burr was born on February 6th, 1756.
Aaron Burr died on September 14th, 1836 at the age of 80 in Port Richmond (Neighborhood on Staten Island). Today Aaron Burr would be 268 years old.
Newark (City in Essex County, New Jersey, United States).
Port Richmond (Neighborhood on Staten Island).
Aaron Burr was born in the zodiac sign Aquarius (The Water-Bearer).
Aaron Burr was born in 1756 in the year of the Chinese zodiac Rat.