How many milliliters of 2.5 M HCl are required
to exactly neutralize 1.5 L of 5.0 M NaOH?

3000mL

Explanation:

The following data were obtained from the question:

Volume of acid (Va) =..?

Molarity of acid (Ma) = 2.5M

Volume of base (Vb) = 1.5L

Molarity of base (Mb) = 5M

Next, we shall write the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

We obtained the following:

Mole ratio of the acid (nA) = 1

Mole ratio of base (nB) = 1

Next, we shall determine the volume of HCl needed for the reaction. This can be obtained as follow illustrated below:

MaVa / MbVb = nA/nB

2.5 x Va / 5 x 1.5 = 1

Cross multiply

2.5 x Va = 5 x 1.5

Divide both side by 2.5

Va = 5 x 1.5 / 2.5

Va = 3L

The volume of the acid required is 3L.

Finally, we shall convert the volume of the acid from litre (L) to millilitre (mL). This is illustrated below:

1L = 1000mL

Therefore, 3L = 3 x 1000 = 3000mL.

Therefore, the volume of the acid, HCl in mL needed fi4 the reaction is 3000mL

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

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