T k = 15 + 273 = 288 K
4.6 / 13 => 0.353 atm
0.50 / 0.10 => 5 L
<span>(15 + 273) K x (13 atm / 7.6 atm) x (0.50 L / 0.10 L)
</span>
<span>= </span>2463.15 K
<span>hope this helps!</span>
Answer:
ecosystem
Explanation:
your welcome hope I helped
Carbon dioxide, water;then energy source in sunlight and chlorophl maybe
Decomp. ~ 2H2O —-> 2H2 + O2
Single Disp. ~ AgNO3 + Cu —> CuNO3 + Ag
Double Disp ~ Pb(NO3)2(aq) + 2KI(aq) -> 2KNO3(aq) + PbI2(s)
Combustion ~ CH4 + O2 —> CO2 + 2H2O
Synthesis ~ 2Mg + O2 —> 2MgO
<u>Answer:</u> The sample of Carbon-14 isotope is 37056.3 years old
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
![t_{1/2}=\frac{0.693}{k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B0.693%7D%7Bk%7D)
where,
= half life of the reaction = 5720 years
Putting values in above equation, we get:
![k=\frac{0.693}{5720yrs}=1.21\times 10^{-4}yrs^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7B5720yrs%7D%3D1.21%5Ctimes%2010%5E%7B-4%7Dyrs%5E%7B-1%7D)
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ![1.21\times 10^{-4}yr^{-1}](https://tex.z-dn.net/?f=1.21%5Ctimes%2010%5E%7B-4%7Dyr%5E%7B-1%7D)
t = time taken for decay process = ? yr
= initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 88.5) = 11.5 grams
Putting values in above equation, we get:
![1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{11.5}\\\\t=37056.3yrs](https://tex.z-dn.net/?f=1.21%5Ctimes%2010%5E%7B-4%7D%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B100%7D%7B11.5%7D%5C%5C%5C%5Ct%3D37056.3yrs)
Hence, the sample of Carbon-14 isotope is 37056.3 years old