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1824 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1824 United States presidential election in Rhode Island

← 1820 October 26 – December 2, 1824 1828 →
 
JohnQAdams.jpg
WilliamHCrawford.jpg
Nominee John Quincy Adams William H. Crawford
Party Democratic-Republican Democratic-Republican
Home state Massachusetts Georgia
Running mate John C. Calhoun Nathaniel Macon
Electoral vote 4 0
Popular vote 2,145 200
Percentage 91.47% 8.53%

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

The 1824 United States presidential election in Rhode Island took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Rhode Island voted for John Quincy Adams over William H. Crawford, Henry Clay, and Andrew Jackson. Adams won Rhode Island by a margin of 82.94%.

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Transcription

Results

1824 United States presidential election in Rhode Island[1]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican John Quincy Adams 2,145 91.47% 4
Democratic-Republican William H. Crawford 200 8.53% 0
Totals 2,345 100.0% 4

References

  1. ^ "1824 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 27 February 2013.


This page was last edited on 10 July 2019, at 13:40
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