Answer:
Rate constant of the reaction is 
.
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
![R=k[A]^a[B]^b[C]^c](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5Ea%5BB%5D%5Eb%5BC%5D%5Ec)
Rate(R) of the reaction in trail 1 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)
![9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[1]
Rate(R) of the reaction in trail 2 ,when :
![[A]=0.30 M,[B]=0.30 M,[C]=0.90 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.30%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.90%20M)
![2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c](https://tex.z-dn.net/?f=2.7%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.90%20M%5D%5Ec)
...[2]
Rate(R) of the reaction in trail 3 ,when :
![[A]=0.60 M,[B]=0.30 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.30%20M%2C%5BC%5D%3D0.30%20M)
![3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=3.6%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.60%20M%5D%5Ea%5B0.30%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[3]
Rate(R) of the reaction in trail 4 ,when :
![[A]=0.60 M,[B]=0.60 M,[C]=0.30 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.60%20M%2C%5BB%5D%3D0.60%20M%2C%5BC%5D%3D0.30%20M)
![3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c](https://tex.z-dn.net/?f=3.6%5Ctimes%2010%5E%7B-4%7D%20M%2Fs%3Dk%5B0.60%20M%5D%5Ea%5B0.60%20M%5D%5Eb%5B0.30%20M%5D%5Ec)
...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
![R=k[A]^2[B]^0[C]^1](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E2%5BB%5D%5E0%5BC%5D%5E1)
Rate constant of the reaction = k
![9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1](https://tex.z-dn.net/?f=9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%3Dk%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1)
![k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B9.0%5Ctimes%2010%5E%7B-5%7D%20M%2Fs%7D%7B%5B0.30%20M%5D%5E2%5B0.30%20M%5D%5E0%5B0.30%20M%5D%5E1%7D)
Dmitri Mendeleev is the scientist who first developed the periodic table.
He is a Russian chemist and inventor who formulated the periodic law and created the periodic table of elements. He corrected some of the properties of the elements. Now, we can easily locate the elements we wanted to identify. With the help of Mendeleev, studying chemistry is not that difficult because he tried to simplify some of the concept that was complex before.
I think answer should be the last option I hope this helps let me know if it’s correct thanks
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
