Answer:
4 g after 58.2 years
0.0156 After 291 years
Explanation:
Given data:
Half-life of strontium-90 = 29.1 years
Initially present: 16g
mass present after 58.2 years =?
Mass present after 291 years =?
Solution:
Formula:
how much mass remains =1/ 2n (original mass) ……… (1)
Where “n” is the number of half lives
to find n
For 58.2 years
n = 58.2 years /29.1 years
n= 2
or 291 years
n = 291 years /29.1 years
n= 10
Put values in equation (1)
Mass after 58.2 years
mass remains =1/ 22 (16g)
mass remains =1/ 4 (16g)
mass remains = 4g
Mass after 58.2 years
mass remains =1/ 210 (16g)
mass remains =1/ 1024 (16g)
mass remains = 0.0156g
Answer: the correct answer is B. Hallogens hope this helped mind helping out with making me brainliest?
<u>Answer:</u> The number of moles of calcium metal required is 2.0 moles.
<u>Explanation:</u>
We are given:
Moles of water reacted = 4.0 moles
The given chemical reaction follows:

By Stoichiometry of the reaction:
2 moles of water is reacted with 1 mole of calcium metal
So, 4.0 moles of water will react with =
of calcium metal
Hence, the number of moles of calcium metal required is 2.0 moles.