B. aluminum is the representative element. :)
Answer:
The rate equation for this reaction:
![R=k[NH_3]^0](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5E0)
Explanation:
Decomposition of ammonia:

Rate law of the can be written as;
![R=k[NH_3]^x](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5Ex)
1) Rate of the reaction , when ![[NH_3]=2.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D2.0%5Ctimes%2010%5E%7B-3%7D%20M)
..[1]
2) Rate of the reaction , when ![[NH_3]=4.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D4.0%5Ctimes%2010%5E%7B-3%7D%20M)
..[2]
[1] ÷ [2]
![\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%7B1.5%5Ctimes%2010%5E%7B-6%7DM%2Fs%7D%3D%5Cfrac%7Bk%5B2.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D%7Bk%5B4.0%5Ctimes%2010%5E%7B-3%7DM%5D%5Ex%7D)
On solving for x , we get ;
x = 0
The rate equation for this reaction:
![R=k[NH_3]^0](https://tex.z-dn.net/?f=R%3Dk%5BNH_3%5D%5E0)
Answer:
180º
Explanation:
Carbon has 4 valence electrons and to complete the octet requires 4 more , so it will share with the oxigen atoms 4 more, forming a single and a pi bond similar to CO2
Since the carbon does not have lone pairs, the geometry around this central atom will be linear and the angle will be 180 , S = C = S
I’m confused what are you asking exactly?
Answer:
4600s
Explanation:

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)
