United States presidential election in Rhode Island, 1836
Main article: United States presidential election, 1836
| Elections in Rhode Island | |||||||||
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The 1836 United States presidential election in Rhode Island took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island voted for Democratic candidate Martin Van Buren over Whig candidate William Henry Harrison. Van Buren won Rhode Island by a margin of 4.48%.
Results
| United States presidential election in Rhode Island, 1836[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 2,964 | 52.24% | 4 | 100.00% | ||
| Whig | William Henry Harrison of Ohio | Francis Granger of New York | 2,710 | 47.76% | 0 | 0.00% | ||
| Total | 5,674 | 100.00% | 4 | 100.00% | ||||
References
- ↑ "1836 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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