SAN FRANCISCO CAHUACUÁ

Oaxaca · MEXICO (MX) · 16.90°N / 97.31°W

ADM1 · ID GeoNames 3,519,286

San Francisco Cahuacuá is a city in Mexico with a population of 503. It sits at 16.90/-97.31, runs on America/Mexico_City, is 12 km from Sindihui.

Population 503

Place Type Populated Place

Elevation n/a

Timezone America/Mexico_City

All facts

fields

Country: Mexico
Subdivision: Oaxaca
Coords: 16.90°N / 97.31°W
UTC offset: -6
DST: Observed

Proximity

nearest 10 · within 50 km

N Sindihui Oaxaca 1,770 12.9 km

N Yutanduchi de Guerrero Oaxaca 1,148 16.0 km

E San Antonio Huitepec Oaxaca 1,594 17.3 km

SW Santa Catarina Cuanana Oaxaca 685 19.8 km

W Santa Cruz Tacahua Oaxaca 578 20.0 km

SW Yutanino Oaxaca 958 20.2 km

S Río Humo Oaxaca 524 20.4 km

S Santiago Textitlán Oaxaca 850 23.1 km

NE Llano de la Canoa Oaxaca 748 23.8 km

E Santiago Huaxolotipac Oaxaca 1,110 24.9 km

FAQ

3 questions

What is the population of San Francisco Cahuacuá?: 503 people, per GeoNames.
Which timezone is San Francisco Cahuacuá in?: America/Mexico_City.
Which country is San Francisco Cahuacuá in?: Mexico (MX).

Get this page as JSON

no auth

cURL 200 OK

$ curl https://api.cityapi.org/v1/cities/3519286

{
  "data": {
    "id": 3519286,
    "name": "San Francisco Cahuacuá",
    "ascii_name": "San Francisco Cahuacua",
    "country_code": "MX",
    "population": 503,
    "elevation": null,
    "timezone_id": "America/Mexico_City",
    "feature_class": "P",
    "feature_code": "PPL",
    "latitude": 16.89619,
    "longitude": -97.30578
  }
}