Answer:
2835 J
Explanation:
Take the specific heat capacity of water as 4.2 J/ g°C.
Energy (heat) = mass x specific heat capacity x change in temperature
(E= mcΔT)
E = 27 x 4.2 x (45-20)
E = 2835 J
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y
3y = 1.2
y = 0,4M
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4
C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄
mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol
164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
Answer:
0.23 V.
Explanation:
<em>∵ ΔG° = -RT lnK.</em>
∴ ΔG° = -RTlnK = -(8.314 J/mol)(298 K) ln(7.3 × 10⁷) = - 44.86 x 10³ J/mol.
<em>∵ ΔG° = - nFE°</em>
∴ E° = - ΔG°/nF = - (- 44.86 x 10³ J/mol)/(2 x 96500 s.A/mol) = 0.2324 V ≅ 0.23 V.
