The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
The First Ionization energy of Nitrogen is greater (Not smaller)than that of Phosphorous. This is because going down the group (N and P are in same group) the number of shells increases, the distance of valence electrons from Nucleus increases and hence due to less interaction between nucleus and valence electrons it becomes easy to knock out the electron.
<span>The second ionization energy of Na is larger than that of Mg because after first loss of electron Na has gained Noble Gas Configuration (Stable Configuration) and now requires greater energy to loose both second electron and Noble Gas Configuration. While Mg after second ionization attains Noble Gas Configuration hence it prices less energy.</span>
Answer:
The time required for the coating is 105 s
Explanation:
Zinc undergoes reduction reaction and absorbs two (2) electron ions.
The expression for the mass change at electrode 
is given as :
where;
M = molar mass
Z = ions charge at electrodes
F = Faraday's constant
I = current
A = area
t = time
also; = 
; replacing that into above equation; we have:
---- equation (1)
where;
A = area
d = thickness
= density
From the above equation (1); The time required for coating can be calculated as;
![[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t](https://tex.z-dn.net/?f=%5B%20%5Cfrac%7B20%20cm%5E2%20%2A0.0025%20cm%2A7.13g%2Fcm%5E3%7D%7B65.38g%2Fmol%7D%2A2%20%5Cfrac%7Bmoles%5C%20of%20%5C%20electrons%7D%7Bmole%20%5C%20of%20%5C%20Zn%7D%20%2A%209.65%2A10%5E4%20%5Cfrac%7BC%7D%7Bmole%20%5C%20of%20%5C%20electrons%20%7D%20%20%5D%20%3D%20%2820%20A%29%20t)
= 105 s
Answer:
Option B
Explanation:
We will check the solubility graph for potassium nitrate, KNO
3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.
Hence, option B is correct
