Abraham O. Smoot American Mormon pioneer
Abraham O. Smoot is a 80 years old American Mormon pioneer from Owenton, Kentucky. Abraham O. Smoot was born on February 17, 1815 (died on March 22, 1895, he was 80 years old) in Owenton, Kentucky as Abraham Owen Smoot.
About
Birth name of Abraham O. Smoot is Abraham Owen Smoot
No, he died on 03/22/1895, 129 years ago. He was 80 years old when he died. He died in Provo
Family
Children
He had three children, Reed (79, United States Senator from Utah) , Brigham (77, American Mormon missionary) and Ida (81) . When his first child, Reed Smoot, was born, Abraham O. Smoot was 46 years old.
What was Abraham's zodiac sign?Abraham O. Smoot zodiac sign was aquarius.
Other facts about Abraham O. Smoot
| Birth name | Abraham Owen Smoot |
|---|---|
| Place of birth | Owenton |
| Place of death | Provo |
| Given name | Abraham |
| Date of death | 1895-03-22T00:00:00Z |
| Religion | The Church Of Jesus Christ Of Latter-day Saints |
| Country of citizenship | United States Of America |
