<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 5730 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 = (100 - 25) = 75 grams
Putting values in above equation, we get:
Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
The answer to this question would be: 2 mol
To answer this question, you need to know the molecular weight of Potassium. Molecular weight determines how much the weight of 1 mol of a molecule has.
Potassium or Kalium molecular weight is 39.1 gram/mol. Then, 78.20gram of potassium should be: 78.20g/ (39.1g/mol)= 2 mol
Any buffer exists in this equilibrium
HA <=>
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (
)
When a strong acid is added, it reacts with the large reservoir of the conjugate base (
) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
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