How can I open a file with its correct program (e.g. ".blend" with Blender and ".webloc" with Chrome) inside a Python program? - python-3.x

Question

I'm working with IDLE on Mac and I'm trying to make a small program that opens a random file from a folder (it is actually a test for a bigger project). In the folder, I have many types of files like ".blend", ".m4a", ".py" and ".webloc" but I expect to have even more in the future. I would like my code to open a random one with their respective program (Blender, QuickTime Player, IDLE, Chrome...) but so far I have not found any way to do it. Is it possible? The most I've been able to do is open Google Chrome from my Windows computer. It doesn't work on my Mac (maybe because it is .app instead of .exe?) and I can only open programs, but not files. Here's the code I used for that:
import subprocess
subprocess.Popen(['C:\Program Files (x86)\Google\Chrome\Application\\chrome.exe', '-new-tab'])
When I enter that on Mac (but with the correct file path for Mac):
import subprocess
subprocess.Popen(['/Applications/Google Chrome.app', '-new-tab'])
It gives me this error (could it be because the file path is written incorrectly? I copied it with right-click on the Chrome file and clicking "copy as path"):
>>>
=============== RESTART: /Users/jaimewalter/Desktop/Test/Test3.py ==============
Traceback (most recent call last):
File "/Users/jaimewalter/Desktop/Test/Test3.py", line 3, in <module>
subprocess.Popen(['/Applications/Google Chrome.app', '-new-tab'])
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/subprocess.py", line 854, in __init__
self._execute_child(args, executable, preexec_fn, close_fds,
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/subprocess.py", line 1702, in _execute_child
raise child_exception_type(errno_num, err_msg, err_filename)
PermissionError: [Errno 13] Permission denied: '/Applications/Google Chrome.app'
>>>
And here's my code for the random file selector:
import random
files = ["Test1.blend", "Test2.m4a", "Test3.py", "Test4.webloc"]
open_this = random.choice (files)
print(open_this)
if open_this == "Test1.blend":
print("Opening Test1.blend")
#now it should open Test1.blend on a new Blender window (/Users/jaimewalter/Desktop/Test/Test1.blend)
elif open_this == "Test2.m4a":
print("Opening Test2.m4a")
#now it should open Test2.m4a on a new QuickTime Player window (/Users/jaimewalter/Desktop/Test/Test2.m4a)
elif open_this == "Test3.py":
print("Opening Test3.py")
#now it should open Test3.py on a new IDLE window or preferably runs the code inside directly it if that's possible (/Users/jaimewalter/Desktop/Test/Test3.py)
elif open_this == "Test4.webloc":
print("Opening Test4.webloc")
#now it should open Test4.webloc on a new Chrome or Safari window (/Users/jaimewalter/Desktop/Test/Test4.webloc)
What should I use to open the files inside the code? Thanks in advance

Answer 1

I already solved it. I used
import subprocess
subprocess.call(["open", "Test1.blend"])
I think with Linux it's xdg-open