Complete and balance the molecular equation for the reaction of aqueous ammonium bromide, NH4BrNH4Br , and aqueous lead(II) acet
1 answer:
<u>Answer: </u>The molecular equation is given below.
<u>Explanation:</u>
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.
This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.
The balanced chemical equation for the reaction of ammonium bromide and lead (II) acetate follows:
This is a type of double displacement reaction because here, exchange of ions takes place.
Hence, the molecular equation is given below.
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Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as
Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as
So time t is given as
So the concentration will become 15% of the initial value after 10.31 hrs.
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
= 0.001 
The change in Concentration Δ![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001
CH3COOH H+ CH3COOH
Initial 
0 0
Change -0.001 +0.001 +0.001
Equilibrium - 0.001 0.001 0.001
Since the
value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
The change in Concentration Δ
CH3COOH H+ CH3COOH
Initial
Change
Equilibrium
Since the
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!