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:
The answer is D, I would type it out but it is to long and I'm kind of lazy, but anyways! Your answer is D
Hope This Helps!
This is possible because of the emulsifying properties present in soap. This property is caused by the hydrophilic end and hydrophobic end of a soap molecule. Grease is able to be dissolved in the water because it is attracted to the hydrophobic end of the soap molecule.
<span><span>N2</span><span>O5</span></span>
Explanation!
When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with
<span><span>25.92 g N<span>14.01 g N/mol N</span></span>=1.850 mol N</span>
<span><span>74.07 g O<span>16.00 g O/mol O</span></span>=4.629 mol O</span>
The ratio between these is <span>2.502 mol O/mol N</span>, which corresponds closely with <span><span>N2</span><span>O5</span></span>.
The formula will be AI2O3
