Explanation:
The metal probably helps to speed up the reaction rate a/c to collision theory, reactant molecules must collide with a reasonable direction by either weakening bonds in reactant molecules to make them extra reactive or by attaching reactant molecules in the exact direction to react.
This is known as an example of heterogeneous catalysis because the catalyst is solid and the reactants are liquid or gases mixture. In this catalysis, the catalyst is present in a different phase compare to the reactants.
Answer:
answer:-
a long-term increase in the Earth's average temperature
Answer:
in crystalline solids
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Answer:
yeast sorry it took so long
Explanation:
Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:
![CH_3COOH~->~CH_3COO^-~+~H^+](https://tex.z-dn.net/?f=CH_3COOH~-%3E~CH_3COO%5E-~%2B~H%5E%2B)
We have an acid (
) and a base (
). Therefore we can write the <u>henderson-hasselbach reaction</u>:
![pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH~%3D~pKa%2BLog%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
If we want to calculate the pH, we have to <u>calculate the pKa</u>:
![pH=-Log~Ka=4.76](https://tex.z-dn.net/?f=pH%3D-Log~Ka%3D4.76)
According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
![[CH_3COO^-]=[CH_3COOH]](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D%5BCH_3COOH%5D)
If we divide:
![\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D~%3D~1)
If we do the Log of 1:
![Log~1=~zero](https://tex.z-dn.net/?f=Log~1%3D~zero)
So:
![pH~=~pKa](https://tex.z-dn.net/?f=pH~%3D~pKa)
With this in mind, the pH is 4.76.
I hope it helps!