B the atmosphere
D. gasoline
C. a carbonated soft drink (without bubbles)
<u>Given:</u>
Initial amount of carbon, A₀ = 16 g
Decay model = 16exp(-0.000121t)
t = 90769076 years
<u>To determine:</u>
the amount of C-14 after 90769076 years
<u>Explanation:</u>
The radioactive decay model can be expressed as:
A = A₀exp(-kt)
where A = concentration of the radioactive species after time t
A₀ = initial concentration
k = decay constant
Based on the given data :
A = 16 * exp(-0.000121*90769076) = 16(0) = 0
Ans: Based on the decay model there will be no C-14 left after 90769076 years
The runner ran a total of 8800 yards.
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>
<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol
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