Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
If the value of H is positive, it means you have to add that much heat to complete the reaction. If H is negative, it means that much heat is released during the chemical process. Because it is -73 kJ, 73 kJ of heat are released in the reaction.
Answer:
Thus, the radius of the helium atom in nanometers is - 0.031 nm
Explanation:
Given that:-
The radius of the helium atom = 31 pm
Considering the conversion of length in pm to the length in nm as:-
1 pm = 0.001 nm
So,
Applying the above conversion factor in the radius of helium atom as:-
Radius = 
nm = 0.031 nm
<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>
Answer:
Option D which is Sn4- is the answer
