The radon-222 sample has a half-life of 3.8 days, and we are asked how many times would the mass divide in half after 23 days. First we calculate the amount of times division occurs by taking the number of days and dividing that by the number of days for one half-life to occur: 23/3.8 = 6.05.
We have 198.6 grams of sample, and we are going to divide it in half 6 times to determine how much of it remains after 23 days:
198.6/2 = 99.3 grams
99.3/2 = 49.65 grams
49.65/2 = 24.83 grams
24.83/2 = 12.41 grams
12.41/2 = 6.21 grams
6.21/2 = 3.1 grams
Therefore, we are left with 3.1 grams of radon-222 after 23 days if one half-life equals to 3.8 days.
Answer:
see explanation
Explanation:
Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction occurs, simply write only NR.
Ag (+1) + NO3(-1) + 2 Na(+1) + Co3 (-2)--> Ag2CO3 (s) + 2 Na (+1) + 2NO3(-1)
The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.
<h3>How do we calculate atomic mass?</h3>
Atomic mass (A) of any atom will be calculated as:
A = mass of protons + mass of neutrons
In the Thorium-234:
Number of protons = 90
Number of neutrons = 144
Mass of one proton = 1.00728 amu
Mass of one neutron = 1.00866 amu
Mass of thorium-234 = 90(1.00728) + 144(1.00866)
Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu
Given mass of thorium-234 = 234.04360 amu
Mass defect = 235.90224 - 234.04360 = 1.85864 amu
Hence required value is 1.85864 amu.
To know more about Atomic mass (A), visit the below link:
brainly.com/question/801533
Because it is a mix of H2O (water) and NaCl (salt)
