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
is 100 then concentration at time t
will be 100-75=25.
Putting the values,

On rearranging,

Thus, time required for 75% decomposition is 21 min.
Answer:
A metalloid is a substance that has both the qualities of a metal and a non metal. Metals conduct electricity and have a high melting point.The element described does not conduct electricity until temperatures are reduced therefore it is not a metal. The fact that conduction occurs when temperatures are reduced means that it is not a non metal because non metals do not conduct electricity at all. Therefore the element is a metaloid because it exhibits some properties of a metal and others of a non mental.
Explanation:
Answer: v2=331.289mL
Explanation:
Formula for ideal gas law is p1v1/T1=p2v2/T2
P1=782.3mmHg
P2=769mmHg at STP
V1=362.4mL
V2=?
T1=273+34.4=307.4k
T2=273k at STP
Then apply the formula and make v2 the subject of formula
V2= 782.3×362.4×273/760×307.4
V2=77397006.96/233624
V2=331.289mL
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V
In a neutral atom they are both equal, and their even quantities makes the atom neutral...