No, you
cannot open "[domain]:8080/[folder]". Tomcat is
not a file server, it is a
web server. What you are "opening" is actually "[domain]:8080/[context]", where "context" is the webapp context root that the webapp has been deployed under. By default, this will be the folder name under TOMCAT_HOME/webapps that you placed your exploded web application (WAR) in - or, if you deplay a WAR file, the name of the WAR file itself (minus the ".war" extension). However, that is the
default, and is often overridden.
Yes, I'm being pedantic, but if you don't understand the critical difference between an application URL context path and a filesystem directory path, you will suffer a lot of confusion. URL resource paths and filesystem paths look very much alike, but they refer to quite different things.
To get a server to pull up a Tomcat web application by simple name using Apache as a reverse proxy, you have to configure Apache to use mod_proxy or mod_jk. Either one works, but I prefer mod_proxy myself - it's the generally recommended one. What they do is set up an internal network tunnel from Apache to Tomcat's port 8009. You can define the proxy either as a context path within Apache (
http://myserver.com/tomcat_app) or as a virtual host (
http://tomcat_app.myserver.com), whichever you like better. Setting up a virtual host does require that you provide doman name resolution for the virtual hostname, but other than that, it's no big deal.
The proxy tunnerl redefines the URL. If a request comes in to "tomcat_app.myserver.com/", and I've defined my proxy link in my Apache VirtualHost to connect to localhost:8009/my_webapp, then that's really all that's needed.