Answer:
![-r_{A}=k\times[BF_3]^{1}\times[NH_3]^{1}](https://tex.z-dn.net/?f=-r_%7BA%7D%3Dk%5Ctimes%5BBF_3%5D%5E%7B1%7D%5Ctimes%5BNH_3%5D%5E%7B1%7D)
Explanation:
The rate law of a chemical reaction is given by
![-r_{A}=k\times[BF_3]^{\alpha}\times[NH_3]^{\beta}](https://tex.z-dn.net/?f=-r_%7BA%7D%3Dk%5Ctimes%5BBF_3%5D%5E%7B%5Calpha%7D%5Ctimes%5BNH_3%5D%5E%7B%5Cbeta%7D)
This law can be written for any experiment, and making the quotient between those expressions the reaction orders can be found
Between experiments 1 and 2
![\frac{-r_{A1}}{{-r}_{A2}}=\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)^\beta](https://tex.z-dn.net/?f=%5Cfrac%7B-r_%7BA1%7D%7D%7B%7B-r%7D_%7BA2%7D%7D%3D%5Cleft%28%5Cfrac%7B%5Cleft%5BNH_3%5Cright%5D_1%7D%7B%5Cleft%5BNH_3%5Cright%5D_2%7D%5Cright%29%5E%5Cbeta)
Then the expression for the calculation of 
![\beta=\frac{ln\frac{-r_{A1}}{-r_{A2}}}{ln\left(\frac{\left[NH_3\right]_1}{\left[NH_3\right]_2}\right)}=\frac{ln\frac{0.2130}{0.1065}}{ln\left(\frac{0.250}{0.125}\right)}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cfrac%7Bln%5Cfrac%7B-r_%7BA1%7D%7D%7B-r_%7BA2%7D%7D%7D%7Bln%5Cleft%28%5Cfrac%7B%5Cleft%5BNH_3%5Cright%5D_1%7D%7B%5Cleft%5BNH_3%5Cright%5D_2%7D%5Cright%29%7D%3D%5Cfrac%7Bln%5Cfrac%7B0.2130%7D%7B0.1065%7D%7D%7Bln%5Cleft%28%5Cfrac%7B0.250%7D%7B0.125%7D%5Cright%29%7D)
Resolving
Doing the same between experiments 3 and 4 the expression for 
is
![\alpha=\frac{ln\frac{-r_{A3}}{-r_{A4}}}{ln\left(\frac{\left[BF_3\right]_3}{\left[BF_3\right]_4}\right)}=\frac{ln\frac{0.0682}{0.1193}}{ln\left(\frac{0.200}{0.350}\right)}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7Bln%5Cfrac%7B-r_%7BA3%7D%7D%7B-r_%7BA4%7D%7D%7D%7Bln%5Cleft%28%5Cfrac%7B%5Cleft%5BBF_3%5Cright%5D_3%7D%7B%5Cleft%5BBF_3%5Cright%5D_4%7D%5Cright%29%7D%3D%5Cfrac%7Bln%5Cfrac%7B0.0682%7D%7B0.1193%7D%7D%7Bln%5Cleft%28%5Cfrac%7B0.200%7D%7B0.350%7D%5Cright%29%7D)
Resolving
This means that the rate law for this reaction is
Answer:
pOH = 1.3, pH = 12.7
Explanation:
Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.
NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M
pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:
pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3
Since pH + pOH = 14, we can plug in pOH and solve for pH:
pH + 1.3 = 14
pH = 14 - 1.3 = 12.7
Thus, our pOH = 1.3 and pH = 12.7.
Isn't it a because in b at the start of the equation the E in Fe just disappeared
Answer:
Explanation:
mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20
mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30
mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67
Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057
mole fraction of N₂ = .5357 / 6.2057 = .0877
mole fraction of He = 1.25 / 6.2057 = .20
mole fraction of CH₄ = 3.75 / 6.2057 = .6043
mole fraction of C₂H₆ = .67 / 6.2057 = .108
Partial pressure = total pressure x mole fraction
Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa
Partial pressure of He = 1200 kPa x .20 = 240 kPa
Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa
Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa
