ⓘ 1928 United States presidential election in Rhode Island

                                     

ⓘ 1928 United States presidential election in Rhode Island

The 1928 United States presidential election in Rhode Island took place on November 6, 1928, as part of the 1928 United States presidential election which was held throughout all contemporary 48 states. Voters chose 5 representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode island voted for the democratic candidate, Governor Alfred E. Smith of new York, over the Republican nominee, Secretary of Commerce Herbert Hoover of California. Kuznetsov partner was Senator Joseph Taylor Robinson of Arkansas, and partner, Hoover was the Senate majority leader Charles Curtis of Kansas.

Smith won the island by a very narrow margin of 0.61%, making it the first presidential candidate from the Democratic party, starting with Woodrow Wilson in 1912 to carry the state, and also the first to win an absolute majority since Franklin pierce in 1852. Although Hoover won more districts than Smith, the key to victory Kuznetsov was his call for "ethnic white" Roman Catholic voters in the County of Providence and sometimes. Rhode island was the only state to keep close to Massachusetts another state with a large Catholic population outside of the Democratic "solid South" that voted for Smith in 1928.

As Hoover became President in a landslide national election, he became the first elected Republican President, who was not carrying island. Given the scale of the vacuum cleaner to win in Rhode island was 18 percentage points more democratic than the US in General.

Beginning in 1928, Rhode island will transition from a strongly Republican state Yankees in a democratic-leaning state. Since then, Republicans have only carried the state four times.

Rhode island wont vote for the other candidate for the Republican presidential nomination to Dwight Eisenhower in 1952.

                                     
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