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The 1819 Rhode Island gubernatorial election was an uncontested election held on April 21, 1819 to elect the Governor of Rhode Island. Nehemiah Rice Knight, the incumbent governor and Democratic-Republican nominee, was the only candidate and so won with 100% of the vote.
General election
Elections in Rhode Island |
---|
Candidates
- Nehemiah Rice Knight, the incumbent governor since 1817.[2]
Results
Party | Candidate | Votes | % | ±% | |
---|---|---|---|---|---|
Democratic-Republican | Nehemiah Rice Knight (incumbent) | 2,670 | 100% | ||
Majority | 2,670 | 100% | |||
Democratic-Republican hold | Swing | ||||
County results
County[3] | Nehemiah Knight
Democratic-Republican |
Total votes | |
---|---|---|---|
# | % | ||
Bristol | 175 | 100% | 175 |
Kent | 473 | 100% | 473 |
Newport | 518 | 100% | 518 |
Providence | 1,056 | 100% | 1,056 |
Washington | 448 | 100% | 448 |
Totals | 2,670 | 100% | 2,670 |
References
- ↑ "Election". The Rhode-Island Republican. Newport, R.I. April 21, 1819. p. 2. Retrieved October 24, 2021.
- ↑ "KNIGHT, Nehemiah Rice". Biographical Directory of the United States Congress. Retrieved June 2, 2022.
- 1 2 "Rhode Island 1819 Governor". A New Nation Votes: American Election Returns 1787-1825. January 11, 2012. Retrieved June 2, 2022.
- ↑ Dubin, Michael J. (2003). United States Gubernatorial Elections, 1776-1860: The Official Results by State and County. Jefferson: McFarland & Company. p. 232. ISBN 9780786414390.
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