1800 United States presidential election in Rhode Island
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This article relies largely or entirely on a single source. (May 2022) |
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County Results
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Elections in Rhode Island |
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The 1800 United States presidential election in Rhode Island took place as part of the 1800 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College who voted for president and vice president.
Rhode Island voted for the Federalist candidate, John Adams, over the Democratic-Republican candidate, Thomas Jefferson. Adams won Rhode Island by a margin of 4.3%. All 4 Adams electors received more votes than the 4 Jefferson electors and the electoral vote was all for Adams in Rhode Island.
Results
1800 United States presidential election in Rhode Island [1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Federalist | John Adams (incumbent) | 2,353 | 52.15% | 4 | |
Democratic-Republican | Thomas Jefferson | 2,159 | 47.85% | 0 | |
Totals | 4,512 | 100.0% | 4 |
See also
References
- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved April 25, 2022.
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