How many ml of 0.01 M NaOH are required to neutralize 50 ml of 0.02 M HCl

100 ml

Explanation:

The neutralization reaction is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Since 1 mol NaOH reacts with 1 mol HCl, at the equivalence point of the titration, the number of moles of NaOH is equal to the number of moles of HCl. We calculate the moles as the product of molarity (M) and volume (V):

moles NaOH = moles HCl

M(NaOH) x V(NaOH) = M(HCl) x V(HCl)

Thus, we calculate the volume of NaOH needed to completely neutralice HCl:

V(NaOH) = (M(HCl) x V(HCl)) /M(NaOH)

               = (0.02 M x 50 ml)/ (0.01 M)

               = 100 ml

Therefore, 100 ml of 0.01 M NaOH are required to neutralize 50 ml of 0.02 M HCl.

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.

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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.