Answer:
autotrophs
Explanation:
4180J
(25.0g)(4.184J/g°C)(75°C-35.0°C)
(25.0g)(40.0°C)(4.184J/g°C)
(1.00*10³g°C)(4.184J/g°C) = 4184J
use sig figs:
It kind of is logical so my answer is yes
A) 0 charge
neutron=neutral
