Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
From the calculations, the half life of the material is 6.5 days.
<h3>What is radioactivity?</h3>
The term radioactivity has to do with the spontaneous disintegration of a specie.
Uisng the formula;
N=Noe^-kt
N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq
No = amount initially present = 1.75 x 10^12 Bq
k = rate constant = ?
t = time taken = 55 days
Hence;
4.995 ×10^9 = 1.75 x 10^12e^-55k
4.995 ×10^9/1.75 x 10^12 = e^-55k
2.85 * 10^-3 = e^-55k
ln2.85 * 10^-3 = -55k
k = ln2.85 * 10^-3/-55
k = 0.1066 day-1
Half life = 0.693/ 0.1066 day-1
= 6.5 days
Learn more about radioactivity:brainly.com/question/1770619
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To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles
The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
B) Mg is the alkaline earth metal w/12 protons so following the periodic table to the halogen in the same period is
Cl: Chlorine
C) The Neutral noble has w/ 18 electrons is argon so the metal in the same row is
Na: Sodium
