What volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

9.85mL

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

Ca(OH)2 + 2HNO3 —> Ca(NO3)2 + 2H2O

From the balanced equation above,

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) =?

Mb (Molarity of base) = 0.155 M

Va (volume of the acid) = 28.8 mL

Ma (Molarity of acid) = 0.106 M

Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i.e the base) can be obtain as follow:

MaVa/MbVb = nA/nB

0.106 x 28.8 / 0.155 x Vb = 2/1

Cross multiply to express in linear form as shown below:

2 x 0.155 x Vb = 0.106 x 28.8

Divide both side by 2 x 0.155

Vb = (0.106 x 28.8) / (2 x 0.155)

Vb = 9.85mL

Therefore, the Volume of calcium hydroxide is 9.85mL

i think   the answer is c hope i'm right

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.