Physical and chemical changes are reversible and nonreversible changes. The release of carbon dioxide gas indicates a chemical change.
<h3>What is a chemical change?</h3>
A chemical change is a change in which a new substance is produced when the reactants react in the reaction mixture. The chemical produced cannot be reversed back to the reactants and hence, it is nonreversible.
The release of carbon dioxide is a chemical change as limestone and hydrochloric acid react with each other chemically and as a result, a new substance was formed.
Therefore, the production of carbon dioxide indicates a chemical change.
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Answer:Yes,enzymes are catalyzed reactions
Explanation:Enzymes are protein that speeds up chemical reactions. Enzyme catalyzed reaction are divided into two:
Homogeneous reaction
Heterogeneous reaction.
Homogeneous catalysts occupy the same phase as the reaction mixture, while heterogeneous catalysts occupy a different phase.
Acid catalysis, organometallic catalysis, and enzymatic catalysis are examples of homogeneous catalysis.
Vanadium oxide (V2 O5) is a brown/yellow solid on which the oxygen and sulfur dioxide can adsorb in order to react with each other to form sulfuric acid.
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
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