A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without itself being consumed by the reaction.
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:


With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

I hope it helps!
Answer:
kinetic energy
Explanation:
When the object is released, the gravitational potential energy is gradually converted into kinetic energy as it picks up speed.
Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is
.
Explanation:
The combined gas equation is,

where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas = 
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:


The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is
.
Answer:
The answer to your question is letter B
Explanation:
Incorrect name
A. acetic acid This name is correct for the acid with formula CH₃COOH
B. hydrocarbonate acid This is not the name for acid but for a molecule that has hydrogen and a metal.
C. hydrocyanic acid This name is correct for the inorganic molecule with formula HCN
D. sulfurous acid This name is correct and is the name of the inorganic molecule with formula H₂SO₃.
E. phosphoric acid This name is correct for the acid with formula H₃PO₄.