<u>Answer:</u> The tree was burned 16846.4 years ago to make the ancient charcoal
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 5715 years
Putting values in above equation, we get:
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 = 
t = time taken for decay process = ? yr
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the sample = 100 grams
[A] = amount left after decay process = 13 grams
Putting values in above equation, we get:
Hence, the tree was burned 16846.4 years ago to make the ancient charcoal
Answer:
69.7% is percent yield
Explanation:
Based on the reaction:
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
2 moles of Na3PO4 react producing 6 moles of NaNO3.
As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:
24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =
72.6 moles of NaNO3
As there are produced 50.6 moles of NaNO3, percent yield is:
50.6 moles NaNO3 / 72.6 moles NaNO3 =
<h3>69.7% is percent yield</h3>
Answer:
b,d
Explanation:
An ion is an atom with a difference in the amount of protons and elektrons.
If a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, then the final volume would be 11,943.8144 ml if the pressure of the gas is changed to 775 torr assuming that the amount and the temperature of the gas remain constant.
It is given that the initial pressure P₁ is 24,650Pa and initial volumeV₁ is 376ml and the final pressureP₂ is 775 torr. We need to find the final volume of the gas. The final volume could be found using the following formula:
P₁V₁ = P₂V₂
By substituting the values, we get
24650 x 376 = 776 x V₂
9268400 = 776V₂
V₂ = 9268400/776
V₂ = 11,943.8144 ml
Therefore, the final volume of the gas would be 11,943.8144 ml
To know more about Partial pressure, click below:
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Answer:
26.981539 u
Atomic number: 13
Symbol: Al
Electron configuration: [Ne] 3s²3p¹
