Let' pretend it is actually 0.035 ohms
Lets be conservative and limit the MOSFET junction temperature to 150ºC (water boils at 100)
Let's also say that the maximum room temperature is 27ºC
The MOSFET junction to ambient thermal resistance is 62ºC/W
So the maximum power dissipation will be (150-27)/62 ~ 2W
Power = I²*R so if your current is 10A, then the max Rds(on) would be 2W/(10²) = 0.02Ω
Thus 0.035Ω is way too much or you need a big heatsink.
The actual max Rds(on) is unknown.