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:
10.08 L.
Explanation:
- If we assume that CO₂ gas behaves ideally at STP (standard T(0.0 °C) and P(1.0 atm)):
<em>It is known that 1.0 mole of ideal gas occupies 22.4 L at STP conditions.</em>
<em></em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of CO₂ gas occupies → 22.4 L.
0.45 mole of CO₂ gas occupies → ??? L.
<em>∴ The volume occupied by 0.45 mole of CO₂ gas </em>= (0.45 mol)(22.4 L)/(1.0 mol) = <em>10.08 L.</em>
Answer:
1.27 atm.
Explanation:
The following data were obtained from the question:
Pressure (P) in torr = 968 torr
Pressure (P) in atm =...?
Thus, we can convert from torr to atm by doing the following:
Recall:
760 torr = 1 atm
Therefore,
968 torr = 968/760 = 1.27 atm
Therefore, 968 torr is equivalent to 1.27 atm.
Answer:
four
Explanation:
adenine, thymine, guanine, and cytosine
