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ALEXANDER, IA 50420

United States » Iowa » Franklin » 50420
The latitude of ALEXANDER, IA 50420 is 42.8198, and the longitude is -93.4294.
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ALEXANDER, IA 50420

ALEXANDER, IA 50420

The latitude of ALEXANDER, IA 50420 is 42.8198, and the longitude is -93.4294.
  • Latitude: 42.8198
  • Longitude: -93.4294
  • State: Iowa
  • Country: United States
  • Continent: North America
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