Albert C. Howard
(1828-1910)

American politician – Albert C. Howard was born in Cranston (City in Providence County, Rhode Island, United States) on February 29th, 1828 and died in on July 3rd, 1910 at the age of 82. Today Albert C. Howard would be 196 years old.

Age

how old was Albert C. Howard when he died?

82

Albert C. Howard died in 1910 at the age of 82.


Biographical data

Birthday February 29th, 1828 (Friday)
Place of Birth Cranston
City in Providence County, Rhode Island, United States
Death Date July 3rd, 1910 (Sunday)
Birth sign (Zodiac) Pisces (The Fish) ♓
Chinese Zodiac Rat 鼠

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Also born in 1828

Albert C. Howard

Birthday

When was Albert C. Howard born?

Albert C. Howard was born on February 29th, 1828.

Death

When did Albert C. Howard die?

Albert C. Howard died on July 3rd, 1910 at the age of 82 in . Today Albert C. Howard would be 196 years old.

Place of Birth

Where was Albert C. Howard born?

Cranston (City in Providence County, Rhode Island, United States).

Zodiac

What is the zodiac sign of Albert C. Howard?

Albert C. Howard was born in the zodiac sign Pisces (The Fish).

Chinese Zodiac

Which Chinese zodiac sign does Albert C. Howard have?

Albert C. Howard was born in 1828 in the year of the Chinese zodiac Rat.

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Died on July 3rd, 1910:
Anniversaries of death

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