Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get


Now we have to calculate the half-life.



Therefore, the half life of 28-Mg in hours is, 6.94
Answer:
Iodine
Explanation:
It's in the same group as chlorine.
Answer:
what do u mean????? i dont get the question
Answer: 281 hours
Explanation:-
1 electron carry charge=
1 mole of electrons contain=
electrons
Thus 1 mole of electrons carry charge=

of electricity deposits 1 mole or 63.5 g of copper
0.0635 kg of copper is deposited by 193000 Coloumb
11.5 kg of copper is deposited by=
Coloumb

where Q= quantity of electricity in coloumbs = 34952756 C
I = current in amperes = 34.5 A
t= time in seconds = ?


Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.