Answer:
The International Date Line, established in 1884, passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich, England, in 1852.
The overall reaction is given by:

The fast step reaction is given as:

The slow step reaction is given as:
(slow step
)
Now, the expression for the rate of reaction of fast reaction is:
![r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=r_%7B1%7D%3Dk_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D-k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
The expression for the rate of reaction of slow reaction is:
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as
takes place in this reaction.
The expression of rate of formation is:

=
(1)
Now, consider that the fast step is always is in equilibrium. Therefore, 
![k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=k_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D%3D%20k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
![[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D%20%3D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D)
Substitute the value of
in equation (1), we get:
![\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28NOBr%29%7D%7Bdt%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
=![k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D%5BNO%5D)
= ![\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7B1%7Dk_%7B2%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D)
Thus, rate law of formation of
in terms of reactants is given by
.
Answer:
B
Explanation:
Noble gases are in group 18 (neon, argon, etc)
You may find bellow the balanced chemical reactions.
Explanation:
To balance chemicals equation the number of atoms of each element
entering the reaction have to be equal to the number of atoms of each
element leaving the reaction, in order to conserve the mass.
2. Single displacement: Zn + 2 HCl → ZnCl₂ + H₂
3. Decomposition: NiCl₂ → Ni + Cl₂
4. Double displacement: 2 LiI + Pb(NO₃)₂ → PbI₂ + 2 LiNO₃
5. Double displacement: BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl
6. Single displacement: Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag
7. Synthesis (combination): 2 Na + O₂ → 2 Na₂O
8. Solid mercury (II) oxide decomposes to form liquid mercury and oxygen.
2 HgO (s) → 2 Hg (l) + O₂ (g) (decomposition reaction)
9. A solution of zinc hydroxide reaction with phosphoric acid (H₃PO₄) to produce solid zinc phosphate and water.
3 Zn(OH)₂ + 2 H₃PO₄ → Zn₃(PO₄)₂ (s) + 6 H₂O (l) (double displacement reaction)
where:
(s) - solid
(l) - liquid
Learn more about:
types of chemical reactions
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in form of solid form honey possess