Explanation:
1) Initial mass of the Cesium-137=
= 180 mg
Mass of Cesium after time t = N
Formula used :
Half life of the cesium-137 = ![t_{1/2}=30 years[p/tex]where, [tex]N_o](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D30%20years%5Bp%2Ftex%5D%3C%2Fp%3E%3Cp%3Ewhere%2C%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DN_o)
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant
Now put all the given values in this formula, we get
Mass that remains after t years.
Therefore, the parent isotope remain after one half life will be, 100 grams.
2)
t = 70 years
N = 35.73 mg
35.73 mg of cesium-137 will remain after 70 years.
3)
N = 1 mg
t = ?
t = 224.80 years ≈ 225 years
After 225 years only 1 mg of cesium-137 will remain.
The answer is more protons than electrons.
Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ
