1824 United States presidential election in Rhode Island

1824 United States presidential election in Rhode Island

← 1820 October 26 – December 2, 1824 1828 →

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%

County Results Adams   80-90%   90-100%

President before election James Monroe Democratic-Republican Elected President John Quincy Adams Democratic-Republican

Elections in Rhode Island
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Ballot measures 2020 Question 1
Providence Mayoral elections 1998 2002 2006 2010 2014 2018 2022
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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 4 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%.

Results

See also

• United States presidential elections in Rhode Island

References

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