"LEO the lion goes GER" means based upon the oxidation and reduction reactions.
<h3>What is oxidation and reduction reaction?</h3>
Oxidation reactions are those reactions in which lossing of electrons are taking place from the substrate molecule and in reduction process gaining of electrons is done by substrate molecules.
The phrase "LEO the lion goes GER" is used in the context of oxidation and reduction reactions. Meaning of the terms:
LEO → Lose electrons = oxidation
GER → Gain electrons = reduction
Hence the given line is based on oxidation and reduction.
To know more about oxidation and reduction, visit the below link:
brainly.com/question/4222605
#SPJ1
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:
D
Explanation:
When lead ions and sulfate ions bond, they form sediment so neither a nor b can be the answer.
The important thing is that two nitrate ions were originally bonded with one lead ion, while two potassium ions bonded with a sulfate ion.
Finally, since potassium and nitrate ions don't form sediment these two ions must remain. Therefore the answer is D
Answer: n∗R=22+273.15/4.2∗5n
P2=n∗R∗T2/V2=n∗R∗33.6+273.15/10
Explanation:
Answer:
A. The balloons will increase to twice their original volume.
Explanation:
Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:
P ∝ 1/V
P = k/V
PV = k (constant)
P = pressure, V = volume.
Let the initial pressure of the balloon be P, i.e. 
, initial volume be V, i.e. 
. The pressure is then halved, i.e. 
Therefore the balloon volume will increase to twice their original volume.
