In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol 
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is 
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration ![[A_{0} ]](https://tex.z-dn.net/?f=%5BA_%7B0%7D%20%5D)
is 100 then concentration at time t ![[A_{t} ]](https://tex.z-dn.net/?f=%5BA_%7Bt%7D%20%5D)
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.
Molar mass CH4 = 16.0 g/mol
* number of moles:
932.3 / 16 => 58.26875 moles
T = 136.2 K
V = 0.560 L
P = ?
R = 0.082
Use the clapeyron equation:
P x V = n x R x T
P x 0.560 = 58.26875 x 0.082 x 136.2
P x 0.560 = 650.76
P = 650.76 / 0.560
P = 1162.07 atm
Answer:
What do you need please to understand?
Answer:
B) K⁺, Sr²⁺ , O²⁻
Explanation:
Potassium is present in group one. It is alkali metal and have one valance electron.Potassium need to lose its one valance electron and form cation to get complete octet.
That's why it shows K⁺.
Sr is alkaline earth metal. It is present in group two. It has two valance electrons. Strontium needed to lose its two valance electrons and get stable electronic configuration.
When it loses its two valance electrons it shows cation with charge of +2.
Sr²⁺
Oxygen is present in group 16. It has sex valance electrons. It needed two more electrons to complete the octet. That's why oxygen gain two electron and form anion with a charge of -2.
O²⁻
