Answer :
(1) The initial rate when [A] is halved and [B] is tripled is, 0.288 M/s
(2) The initial rate when [A] is tripled and [B] is halved is, 0.096 M/s
Explanation:
The given rate law expression is:
![Rate=k[A][B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5BB%5D%5E2)
Now we have to determine the initial rate when [A] is halved and [B] is tripled.
The new rate law expression will be:
![Rate=k\times (\frac{[A]}{2})\times (3\times [B])^2](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B%5BA%5D%7D%7B2%7D%29%5Ctimes%20%283%5Ctimes%20%5BB%5D%29%5E2)
![Rate=k\times (\frac{[A]}{2})\times 9\times [B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B%5BA%5D%7D%7B2%7D%29%5Ctimes%209%5Ctimes%20%5BB%5D%5E2)
![Rate=k\times (\frac{9}{2})\times [A]\times [B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B9%7D%7B2%7D%29%5Ctimes%20%5BA%5D%5Ctimes%20%5BB%5D%5E2)
Given:
Initial rate = 0.0640 M/s
As, Initial rate =
= 0.0640 M/s
Thus,


Now we have to determine the initial rate when [A] is tripled and [B] is halved.
The new rate law expression will be:
![Rate=k\times (\frac{[B]}{2})^2\times (3\times [A])](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B%5BB%5D%7D%7B2%7D%29%5E2%5Ctimes%20%283%5Ctimes%20%5BA%5D%29)
![Rate=k\times (\frac{[B]^2}{4})\times 3\times [A]](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B%5BB%5D%5E2%7D%7B4%7D%29%5Ctimes%203%5Ctimes%20%5BA%5D)
![Rate=k\times (\frac{3}{4})\times [A]\times [B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5Ctimes%20%28%5Cfrac%7B3%7D%7B4%7D%29%5Ctimes%20%5BA%5D%5Ctimes%20%5BB%5D%5E2)
Given:
Initial rate = 0.0640 M/s
As, Initial rate =
= 0.0640 M/s
Thus,


Answer:
exothermic reaction
Explanation:
I think this because this is an example of an exothermic reaction is the chemical reaction between sodium and chlorine, which results in the formation of sodium chloride (also known as common salt)
Answer:
d is the right one I think
4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.
If 1 mole of vinegar contains 6.02 x 10^23 particles
x moles of vinegar contains 9.02 x 10^24 particles
x = 1 mole x 9.02 x 10^24 /6.02 x 10^23
x = 15 moles of vinegar
The reaction is as follows;
2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2
Since 2 moles of vinegar reacts with 1 mole of carbonate
x moles of vinegar reacts with 16.5 moles of carbonate
x = 2 moles x 16.5 moles/ 1 mole
x = 33 moles of vinegar
We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.
Theoretical yield = 16.5 moles x 158 g/mol = 2607 g
Actual yield = 6.35 moles x 158 g/mol = 1066.8 g
Percent yield = 1066.8 g/2607 g × 100/1
= 41%
Learn more: brainly.com/question/13440572?