Suppose the concentration of the NaOH solution was 0.5 M instead of 0.1 M. Would this titration have required more, less, or the same amount of NaOH solution for a complete reaction? Explain

The titration will require a less amount of the NaOH solution.

I will like to state here that the amount we are talking about is the volume of the NaOH solution.

Concentration is defined as mole per unit volume i.e

We can see from the formula above that the concentration is inversely proportional to the volume.

This means that a higher concentration requires a lower volume and a lesser concentration requires a higher volume.

Therefore, we can conclude that the titration will require a less amount (i.e volume) of the 0.5 M NaOH solution.

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Less.

Explanation:

Since there was an increase of concentration of NaOH, there would be more molecules of NaOH present. There would be no need to use the same amount of NaOH because there was already more in the solution.

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.