Election

1824 United States presidential election in Rhode Island

election in Rhode Island

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%.

Fetching the article from Wikipedia…

Text from the Wikipedia article “1824 United States presidential election in Rhode Island”, available under CC BY-SA 4.0.

Meanwhile, elsewhere

Send feedback

An idea, a mistake in the history, something that doesn’t work: every message is read.

It is about

Sent with what is on this screen (above), its address and the theme. Your browser, device, languages and rough location are read from the request itself, as any website receives them. How it is kept: privacy.