1920 Iowa Senate election

1920 Iowa Senate election

← 1918 November 2, 1920 1922 →

29 out of 50 seats in the Iowa Senate
26 seats needed for a majority
  Majority party Minority party
 
Party Republican Democratic
Last election 45 5
Seats after 48 2
Seat change Increase3 Decrease3

Results
     Republican gain
     Republican hold

The 1920 Iowa Senate elections took place as part of the biennial 1920 United States elections. Iowa voters elected state senators in 29 of the senate's 50 districts. State senators serve four-year terms in the Iowa Senate.

A statewide map of the 50 state Senate districts in the 1920 elections is provided by the Iowa General Assembly here.

The primary election on June 7, 1920, determined which candidates appeared on the November 2, 1920 general election ballot.[1][2]

Following the previous election, Republicans had control of the Iowa Senate with 45 seats to Democrats' 5 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 21 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1920 general election with the balance of power shifting to Republicans holding 48 seats and Democrats having 2 seats (a net gain of 3 seats for Republicans). Democrats suffered a disastrous performance as they lost every single district up for election, including three held by them.

  1. ^ "Primary Election 1920 For State Senator" (PDF). Iowa General Assembly. Retrieved June 9, 2021.
  2. ^ "General Election 1920 For State Senator" (PDF). Iowa General Assembly. Retrieved June 9, 2021.

1920 Iowa Senate election

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