when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
A. The concentration is in mol/L
Answer: 11, Na, 23, 100, −9.529 ... phosphorus, 15, P, 31, 100, −24.441 ... manganese, 25, Mn, 55, 100, −57.706.
Explanation: Make me Brainelist
Answer: no se wey jaja que le pusiste tu ?
Explanation:
