First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Answer:
ai) Rate law, ![Rate = k [CH_3 Cl] [Cl_2]^{0.5}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BCH_3%20Cl%5D%20%5BCl_2%5D%5E%7B0.5%7D)
aii) Rate constant, k = 1.25
b) Overall order of reaction = 1.5
Explanation:
Equation of Reaction:

If
, the rate of backward reaction is given by:
![Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%5C%5Ck%20%3D%20%5Cfrac%7BRate%7D%7B%20%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%7D%5C%5Ck%20%3D%20%5Cfrac%7BRate%7D%7B%20%5BCH_3%20Cl%5D%5E%7Ba%7D%20%5BCl_2%5D%5E%7Bb%7D%7D)
k is constant for all the stages
Using the information provided in lines 1 and 2 of the table:
![0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1](https://tex.z-dn.net/?f=0.014%20%2F%20%5B0.05%5D%5Ea%20%5B0.05%5D%5Eb%20%3D%2000.029%2F%20%5B0.100%5D%5Ea%20%5B0.05%5D%5Eb%5C%5C0.014%20%2F%20%5B0.05%5D%5Ea%20%5B0.05%5D%5Eb%20%3D%2000.029%2F%20%5B2%2A0.05%5D%5Ea%20%5B0.05%5D%5Eb%5C%5C0.014%20%2F%20%3D%200.029%2F%202%5Ea%5C%5C2%5Ea%20%3D%202.07%5C%5Ca%20%3D%201)
Using the information provided in lines 3 and 4 of the table and insering the value of a:
![0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\](https://tex.z-dn.net/?f=0.041%20%2F%20%5B0.100%5D%5Ea%20%5B0.100%5D%5Eb%20%3D%200.115%20%2F%20%5B0.200%5D%5Ea%20%5B0.200%5D%5Eb%5C%5C0.041%20%2F%20%5B0.100%5D%5Ea%20%5B0.100%5D%5Eb%20%3D%200.115%20%2F%20%5B2%20%2A%200.100%5D%5Ea%20%5B2%20%2A%200.100%5D%5Eb%5C%5C)
![0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5](https://tex.z-dn.net/?f=0.041%20%3D%200.115%20%2F%20%5B2%20%5D%5Ea%20%5B2%5D%5Eb%5C%5C%20%5C%5B%5B2%20%5D%5Ea%20%5B2%5D%5Eb%20%3D%200.115%2F0.041%5C%5C%20%5C%5B%5B2%20%5D%5Ea%20%5B2%5D%5Eb%20%3D%202.80%5C%5C%5C%5B%5B2%20%5D%5E1%20%5B2%5D%5Eb%20%3D%202.80%5C%5C%5C%5B%5B2%5D%5Eb%20%3D%201.40%5C%5Cb%20%3D%20%5Cfrac%7Bln%201.4%7D%7Bln%202%7D%20%5C%5Cb%20%3D%200.5)
The rate law is: ![Rate = k [CH_3 Cl] [Cl_2]^{0.5}](https://tex.z-dn.net/?f=Rate%20%3D%20k%20%5BCH_3%20Cl%5D%20%5BCl_2%5D%5E%7B0.5%7D)
The rate constant
then becomes:
![k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25](https://tex.z-dn.net/?f=k%20%3D%200.014%20%2F%20%28%20%5B0.050%5D%20%5B0.050%5D%5E%280.5%29%20%29%5C%5Ck%20%3D%201.25)
b) Overall order of reaction = a + b
Overall order of reaction = 1 + 0.5
Overall order of reaction = 1.5
boron, carbon, nitrogen, oxygen and flourine families in addition to the noble gases.
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also to answer the 'hello guys what is good!':nothin u?
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The mass of magnesium should be less than 0.09g to enable a faster reaction rate. Magnesium reacts to form a white coating around it which stops the reaction. The lesser the gram the faster the reaction before the coating is formed. It is also advisable to use magnesium fillings to increase the rate of reaction.
Answer:
Option B is correct More hydrogen carbonate is formed
Explanation:
Step 1: Data given
The Principle of Le Chatelier says 'If the concentration of one of the reaction partners changes, the balance will shift to counteract that concentration change.
Step 2: When adding more H2O
For the equation CO2 + H2O --> H2CO3
If the H2O concentration is increased, the system will attempt to undo that change in concentration by shifting the balance to the right, and so the H2CO3 concentration will increase.
Option B is correct More hydrogen carbonate is formed