Answer:
The catalyst speeds up the chemical reaction.
Explanation:
The process in which chemical reaction rate increases by addition of substance is termed as a catalyst. The chemical reaction happens faster when a catalyst is present because it gives an alternative reaction pathway with lower activation energy.
The catalyst usually forms a temporary intermediate that helps in regenerating the original catalyst in the cyclic process. Yeast when added in a mixture work as a catalyst and helps with chemical reaction.
Answer:
B) Conduction requires two objects to be in physical contact.
Explanation:
Conduction is when two objects are directly touching each other. So, the objects must come in physical contact with each other.
Answer:
Niels Bohr, refined the model of an atom by proposing a quantized shell structure atomic model in order to describe how the electrons are able to maintain stable orbits around the nucleus
Based on the predictions of classical mechanics the electron motion of the Rutherford model was unstable as the electrons where expected to have lost some energy during motion and thus having to come rest in the nucleus
According to the modification by Neils Bohr in 1913, electrons move in shells or orbits of fixed energy and emission of electromagnetic radiation takes place only when electrons changes the orbit in which they move
Explanation:
Answer: The excess amount changes from a gas into a liquid or solid (ice). The process of water changing from a gas to a liquid is called "condensation," and when gas changes directly into a solid, it is called "deposition." These two processes are how clouds form.
Answer:

![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
Explanation:
Hello there!
In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:
![\frac{1}{3} =\frac{x}{[Mg(OH)_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)

Best regards!