Determine the volume of 0.150 m naoh solution required to neutralize each sample of hydrochloric acid. the neutralization reaction is: naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) 25 ml of a 0.150 m hcl solution 55 ml of a 0.055 m hcl solution 175 ml of a 0.885 m hcl solution

The Volume of NaOH required to neutralize

1) 25 mL of a 0.150M HCl solution is 25 mL

2) 55 mL of a 0.055M HCl solution is 20.2 mL

3) 175 mL of 0.885M HCl solution is 1032.5 mL

Explanation: Neutralization reactions are defined as the reactions when an acid reacts with a base to form a salt and water.

For the reaction of HCl and NaOH,

To calculate the molarity of NaOH, we use the formula:

1.) 25 mL of a 0.150M of HCl solution.

Putting the values in above equation, we get

2.) 55 mL of a 0.055M of HCl solution.

Putting the values in above equation, we get

3.) 175 mL of a 0.885M of HCl solution.

Putting the values in above equation, we get

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each chemical family is represented as a group or column on the periodic table. based on chemical properties, there are 6 types of chemical families. group 1 represents alkali metals, group 2 represents alkaline earth metals, groups 3 to 12 represent transition metals, groups 13 to 16 represent non-metals and other metals, group 17 represents halogens and group 18 represents noble gases.