Calculate the molarity of a calcium chloride solution if 33.8 mL of it are needed to react with 41.2 mL of 0.877 M AgNO3. Report your answer in standard notation with three significant figures, and do not include units in your answer.

0.465 M IS THE MOLARITY OF CALCIUM CHLORIDE SOLUTION.

Explanation:

Volume of CaCl2 = 33.8 mL = 33.8 / 1000 L = 0.0338 L

Volume of AgNO3 = 41.2 mL = 0.0412 L

Molarity of AgNO3 = 0.877 M

First is to write and balance out the equation for the reaction

2AgNO3 + CaCl2 > 2AgCl + Ca(NO3)2

From the reaction, we could see that 2 moles of AgNO3 combines with 1 mole of CaCl2 to produce 12 moles of AgCl and 1 mole of Ca(NO3)2.

In volmetric analysis, the relationship between molarity and volume of acids and bases is shown in the equation below:

MaVa / Mb Vb = Na / Nb

Ma = molarity of acid = 0.877 M

Va = volume of acid = 0.0412 L

Mb = molarity of base = unknown

Vb = volume of base = 0.0338 L

Na = number of moles of acid = 2

Nb = number of mole of base = 1

Rearranging the formulae, making Mb the subject of the equation, we get:

Mb = Ma Va Nb / Vb Na

Mb = 0.877 * 0.0412 * 1 / 0.0338 * 2

Mb = 0.0361 / 0.0776

Mb = 0.465 M

So therefore, the molarity of calcium chloride solution if the variables given were used is 0.465 M.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

i think so it’s will be b yes i believe its