Part iv. for each trial, calculate the number moles of 6.0 m hcl used in the reaction. report your answer using 4 digits. note that it is 1 or 2 digits beyond the number of significant figures, i. e. 0.2345 moles.

This is an incomplete question, here is a complete question.

For each trial, calculate the number moles of 6.0 M HCl used in the reaction?

Trial 1 : Volume of HCl = 15.0ml

Trial 2 : Volume of HCl = 14.9ml

Trial 3 : Volume of HCl = 15.2ml

Answer :

The number moles of HCl for trial 1, 2 and 3 is, 0.090 mol, 0.089 mol and 0.091 mol

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

In this question, the solute is HCl.

Now we have to calculate the number of moles of HCl for trial 1.

Volume of HCl = 15.0 mL = 0.015 L

Now we have to calculate the number of moles of HCl for trial 2.

Volume of HCl = 14.9 mL = 0.0149 L

Now we have to calculate the number of moles of HCl for trial 3.

Volume of HCl = 15.2 mL = 0.0152 L

i am going to put the answers in order i dint feel like typing it out so yeah

a

d

c

b

b

b

a

d

a

b

c

c

c

b

d

the correct answer is a