1873 Rhode Island gubernatorial election
Appearance
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County results Howard: 60–70% 70–80% >90% | |||||||||||||||||
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Elections in Rhode Island |
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The 1873 Rhode Island gubernatorial election was held on 2 April 1873 in order to elect the governor of Rhode Island. Republican nominee Henry Howard defeated Democratic nominee Benjamin G. Chace.[1]
General election
[edit]On election day, 2 April 1873, Republican nominee Henry Howard won the election by a margin of 5,870 votes against his opponent Democratic nominee Benjamin G. Chace, thereby retaining Republican control over the office of governor. Howard was sworn in as the 32nd governor of Rhode Island on 6 May 1873.[2]
Results
[edit]Party | Candidate | Votes | % | |
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Republican | Henry Howard | 9,656 | 71.68 | |
Democratic | Benjamin G. Chace | 3,786 | 28.11 | |
Scattering | 29 | 0.21 | ||
Total votes | 13,471 | 100.00 | ||
Republican hold |
References
[edit]- ^ "Henry Howard". National Governors Association. Retrieved 8 April 2024.
- ^ "RI Governor". ourcampaigns.com. 6 October 2005. Retrieved 8 April 2024.
Categories:
- Rhode Island gubernatorial elections
- 1873 United States gubernatorial elections
- April 1873 events
- 1873 in Rhode Island
- 1870s in Rhode Island
- 1870s Rhode Island elections
- 1873 elections
- 1873 elections in North America
- 1873 elections in the United States
- United States gubernatorial elections in the 1870s
- Government of Rhode Island