Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>
3.8 /
=
((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
Answer:
So, the right answer is. No. of moles of FeS₂ = 0.25 mole. Explanation: From the balanced. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2.
Explanation:
we have to know the spin of valence electrons of carbon-14
There are four unpaired electron which are called as valence electron also.The spin of the four unpaired electron is either upfilled or down filled.
The ground state electronic configuration of C-atom is 1s²2s²2p² and one electron from 2s orbital gets excited to 2p orbital. The elctronic configuration in excited state is 1s²2s¹
.
The electron jumps because half-filled orbitals are more stable. Exchange energy is less than pairing energy.
Answer: The overall equation will be 
Explanation:
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.
Anode : 
Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.
Cathode : 
The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :
Answer:
7 chlorine atoms
Explanation:
K=2.8.8.1
Cl=2.8.7
pottasium will give chlorine its I valence electron to form ions as follows
K=(2.8.8)+
Cl=(2.8.8)-
