Answer:
T final = 80°C
Explanation:
∴ Q = 18000 cal
∴ m H2O = 300 g
∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K
∴ T1 = 20°C = 293 K
∴ T2 = ?
⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)
⇒ (18000 cal)/(300 cal/K) = T2 - 293 K
⇒ T2 = 293 K + 60 K
⇒ T2 = 353 K (80°C)
<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.
<u>Explanation:</u>
All radioactive decay processes undergoes first order reaction.
To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:
![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%20%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 1.52 hrs
= Initial concentration of reactant = 100 g
[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g
Putting values in above equation, we get:

To calculate the half life period of first order reaction, we use the equation:

where,
= half life period of first order reaction = ?
k = rate constant = 
Putting values in above equation, we get:

Hence, the half life of the sample of silver-112 is 3.303 hours.
Answer:
In first shell 2 electrons are present and 7 electrons are present in last shell
Explanation:
Answer:
B protons determine indentity and valence electrons determine chemical properties.
:)
Explanation: