‼️i will mark brainliest if it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid sulfur the molarity of the nitric acid solution. show all of the work used to solve this problem unbalaced equation: ba(oh)2+hno3-> ba(no3)2+h2o

Ba(OH)2 + 2 HNO3 > Ba(NO3)2 + 2 H2O

1) Calculate moles of Ba(OH)2 used.

0.65 mol / L x 0.0234 L = 0.01521 mol Ba(OH)2

2) Calculate moles of HNO3 used.

0.01521 mol Ba(OH)2 x 2 HNO3 / 1 Ba(OH)2 = 0.03042 mol HNO3

3) Calculate molarity of HNO3.

0.03042 mol / 0.0425 L =

so the answer is 0.716 M !

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