An original sample of K-40 has a mass of 25.00 grams.
After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged.
What is the half-life of K-40?
First step is to determine the remaining decimal amount.
3.125 grams /25.00 grams = 0.125
Second step is to determine the number of half lives.
(1/2)^n = 0.125
N log (1/2) = log 0.125
N = 3 years
Answer:
Explanation:
Considering,
Using ideal gas equation as:
where,
P is the pressure = 760 mmHg
V is the volume = 100.0 mL = 0.1 L
m is the mass of the gas = 0.193 g
M is the molar mass of the gas = ?
Temperature = 17 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (17 + 273.15) K = 290.15 K
R is Gas constant having value = 62.36367 L. mmHg/K. mol
Applying the values as:-
M = 45.95 g/mol
This mass corresponds to . Hence, the gas must be .
Optimization helps you make better choices when you have all the data, and simulation helps you understand the possible outcomes when you don’t. Frontline Solvers enable you to combine these analytic methods, so you can make better choices for decisions you do control, taking into account the range of potential outcomes for factors you don’t control.
Covalent bonds keep the hydrogen and oxygen atoms together
