In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>=

=

= 0.45 mol
Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.
According to the ideal gas equation, we have PV = nRT
V =

.
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V =

= 34.8 l
Answer:
Quantity of
generated will be reduced to fifty percent of its initial amount.
Explanation:
Equilibrium reaction: 
In accordance with the balanced equation, 1 mol of HCl interacts with 1 mol of NaOH leading to the formation of 1 mol of 
<pif the="" quantities="" of="" reactants="" and="" hcl="" are="" diminished="" by="" half="" it="" results="" in:="">
0.5 mol of HCl interacting with 0.5 mol of NaOH yielding 0.5 mol of
.
Consequently, it is clear that the total of
produced will be halved if the quantities of the reactants are halved.
</pif>
Answer:
The rate law for the decomposition reaction is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be 
Explanation:

The rate law can be expressed as:
..[1]
When the drug concentration is tripled, the decomposition rate rises by a factor of nine.
![[D]'=3[D]](https://tex.z-dn.net/?f=%5BD%5D%27%3D3%5BD%5D)

...[2]
[1] ÷ [2]
![\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR%27%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5BD%27%5D%5Ex%7D)
![\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B9R%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5B3D%5D%5Ex%7D)

Solving for x results in:
x = 2.
This indicates a second-order reaction.
The decomposition reaction's rate law is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be:
![k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BR%7D%7B%5BD%5D%5E2%7D%3D%5Cfrac%7BM%2Fs%7D%7B%28M%29%5E2%7D%3DM%5E%7B-1%7Ds%5E%7B-1%7D)
The unit for the rate constant will be
.