The given question is incomplete. The complete question is :
For the net reaction:
, the following slow first steps have been proposed.
A. ![2C+AB\rightarrow AC+BC](https://tex.z-dn.net/?f=2C%2BAB%5Crightarrow%20AC%2BBC)
B. ![2AB\rightarrow A_2+2B](https://tex.z-dn.net/?f=2AB%5Crightarrow%20A_2%2B2B)
C. ![C+AB\rightarrow BC+A](https://tex.z-dn.net/?f=C%2BAB%5Crightarrow%20BC%2BA)
D. ![AB \rightarrow A + B](https://tex.z-dn.net/?f=AB%20%5Crightarrow%20A%20%2B%20B)
Determine the units of the rate constant for all four reactions listed in the problem above, and enter the correct choices from the list below. Enter 4 letters in order (e.g. ABCD or CBED)
a. ![moleL^{-1 }sec^{-1}](https://tex.z-dn.net/?f=moleL%5E%7B-1%20%7Dsec%5E%7B-1%7D)
b. ![mole^{-1}Lsec^{-1}](https://tex.z-dn.net/?f=mole%5E%7B-1%7DLsec%5E%7B-1%7D)
c. ![mole^2L^{-2}sec^{-1}](https://tex.z-dn.net/?f=mole%5E2L%5E%7B-2%7Dsec%5E%7B-1%7D)
d. ![mole^{-2}L^2sec^{-1}](https://tex.z-dn.net/?f=mole%5E%7B-2%7DL%5E2sec%5E%7B-1%7D)
e. None of the above.
Answer: A.
: ![mol^{-2}L^2s^{-1}](https://tex.z-dn.net/?f=mol%5E%7B-2%7DL%5E2s%5E%7B-1%7D)
B.
: ![k=mol^{-1}Ls^{-1}](https://tex.z-dn.net/?f=k%3Dmol%5E%7B-1%7DLs%5E%7B-1%7D)
C.
: ![k=mol^{-1}Ls^{-1}](https://tex.z-dn.net/?f=k%3Dmol%5E%7B-1%7DLs%5E%7B-1%7D)
D.
: ![k=s^{-1}](https://tex.z-dn.net/?f=k%3Ds%5E%7B-1%7D)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
1. ![2C+AB\rightarrow AC+BC](https://tex.z-dn.net/?f=2C%2BAB%5Crightarrow%20AC%2BBC)
![Rate=k[C]^2[AB}^1](https://tex.z-dn.net/?f=Rate%3Dk%5BC%5D%5E2%5BAB%7D%5E1)
![molL^{-1}s^{-1}=k[molL^{-1}]^2[molL^{-1}}^1](https://tex.z-dn.net/?f=molL%5E%7B-1%7Ds%5E%7B-1%7D%3Dk%5BmolL%5E%7B-1%7D%5D%5E2%5BmolL%5E%7B-1%7D%7D%5E1)
![k=mol^{-2}L^2s^{-1}](https://tex.z-dn.net/?f=k%3Dmol%5E%7B-2%7DL%5E2s%5E%7B-1%7D)
2. ![2AB\rightarrow A_2+2B](https://tex.z-dn.net/?f=2AB%5Crightarrow%20A_2%2B2B)
![Rate=k[AB]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BAB%5D%5E2)
![molL^{-1}s^{-1}=k[molL^{-1}]^2](https://tex.z-dn.net/?f=molL%5E%7B-1%7Ds%5E%7B-1%7D%3Dk%5BmolL%5E%7B-1%7D%5D%5E2)
![k=mol^{-1}Ls^{-1}](https://tex.z-dn.net/?f=k%3Dmol%5E%7B-1%7DLs%5E%7B-1%7D)
3. ![C+AB\rightarrow BC+A](https://tex.z-dn.net/?f=C%2BAB%5Crightarrow%20BC%2BA)
![Rate=k[C}^1[AB]^1](https://tex.z-dn.net/?f=Rate%3Dk%5BC%7D%5E1%5BAB%5D%5E1)
![molL^{-1}s^{-1}=k[molL^{-1}]^1[molL^{-1}]^1](https://tex.z-dn.net/?f=molL%5E%7B-1%7Ds%5E%7B-1%7D%3Dk%5BmolL%5E%7B-1%7D%5D%5E1%5BmolL%5E%7B-1%7D%5D%5E1)
![k=mol^{-1}Ls^{-1}](https://tex.z-dn.net/?f=k%3Dmol%5E%7B-1%7DLs%5E%7B-1%7D)
4. ![AB \rightarrow A + B](https://tex.z-dn.net/?f=AB%20%5Crightarrow%20A%20%2B%20B)
![Rate=k[AB]^1](https://tex.z-dn.net/?f=Rate%3Dk%5BAB%5D%5E1)
![molL^{-1}s^{-1}=k[molL^{-1}]^1](https://tex.z-dn.net/?f=molL%5E%7B-1%7Ds%5E%7B-1%7D%3Dk%5BmolL%5E%7B-1%7D%5D%5E1)
![k=s^{-1}](https://tex.z-dn.net/?f=k%3Ds%5E%7B-1%7D)
The volume required of 1.25M HCL to prepare 180ml of a 0.500M Hcl solution is 72ml
It is Balanced but not in its lowest form.
This is because the coefficients could have been reduced to 2,1&3.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of
Answer:
0.5
Explanation:
We are given the moles of two reactants, so this could be a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2Ca₃(PO₄)₂ + 6SiO₂ + 10C → P₄ + 6CaSiO₃ + 10CO
n/mol: 1 3
Calculate the moles of P₄ that can be formed from each reactant
:
1. From Ca₃(PO₄)₂
![\text{Moles of P}_{4} = \text{1 mol Ca$_{3}$(PO}_{4})_{2} \times \dfrac{\text{1 mol P}_{4}}{\text{2 mol Ca$_{3}$(PO}_{4})}_{2} = \text{0.5 mol P}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20P%7D_%7B4%7D%20%3D%20%5Ctext%7B1%20mol%20Ca%24_%7B3%7D%24%28PO%7D_%7B4%7D%29_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20P%7D_%7B4%7D%7D%7B%5Ctext%7B2%20mol%20Ca%24_%7B3%7D%24%28PO%7D_%7B4%7D%29%7D_%7B2%7D%20%3D%20%5Ctext%7B0.5%20mol%20P%7D_%7B4%7D)
2. From SiO₂
![\text{Moles of P}_{4} = \text{3 mol SiO}_{2} \times \dfrac{\text{1 mol P}_{4}}{\text{6 mol SiO}_{2}} = \text{0.5 mol P}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20P%7D_%7B4%7D%20%3D%20%5Ctext%7B3%20mol%20SiO%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20P%7D_%7B4%7D%7D%7B%5Ctext%7B6%20mol%20SiO%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B0.5%20mol%20P%7D_%7B4%7D)
Each reactant forms 0.5 mol of P₄.