I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>
<u>Answer:</u> The correct answer is Option 5.
<u>Explanation:</u>
- To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

where,
are the n-factor, molarity and volume of the NaOH.
are the n-factor, molarity and volume of the 
We are given:
Putting all the values in above equation, we get:

- To calculate the molarity of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base.
We are given:

Putting values in above equation, we get:

Hence, the correct answer is Option 5.
Answer:
True
Explanation:
ion: an atom or molecule with a net electric charge due to the loss or gain of one or more electrons.
According to the illustration, the vanadium (V) oxide would be a catalyst.
<h3>What are catalysts?</h3>
Catalysts are substances that are utilized in reactions that are not themselves consumed in reactions but only speed up the rate of the reactions.
Catalysts speed up the rate of reactions by lowering the activation energy of the reactants.
Sulfur dioxide reacts with oxygen to produce sulfur trioxide. The vanadium (v) oxide is not consumed in the reaction. Thus it only serves as a catalyst.
More on catalysts can be found here: brainly.com/question/12260131
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