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A student added 0.90 mL of 0.50 M NaOH, 0.70 mL of 1.0 M HA, and 0.40 mL water to their trial 1. How many moles of NaOH did they add

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There are 0.00045 moles of sodium hydroxide that were added in the solution.

From the question, we have the following information;

Volume of sodium hydroxide =  0.90 mL or 0.00009 L

Concentration of sodium hydroxide = 0.50 M

We  know that;

Number of moles = volume of solution × concentration of solution

Substituting values;

The number of moles of sodium hydroxide added = 0.9/1000 L ×  0.50

= 0.00045 moles

Learn more: https://brainly.com/question/4612545

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