Answer:
You need to add 400mL of water
Explanation:
500mL = 5 M HCI That means that if you divide both sides by 5
100mL = 1 M HCI If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to
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
Answer:
SAMPLE A - pure substance.
SAMPLE B - homogeneous mixture.
SAMPLE C - heterogeneous mixture.
Explanation: