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Calculate the volume of 1.5 M nitric acid required to neutralize 30.0 mL of 7.0 M sodium
hydroxide.

HNO3 (aq) + NaOH (aq) → H2O (l) + NaNO3 (aq)

Respuesta :

40042

Answer:

0.14 liters nitric acid

Explanation:

Molarity = moles / liter

7.0 M = x / 0.03 L

= 0.21 moles NaOH

0.21 moles NaOH x ( 1 mole HNO3 / 1 mole NaOH) = 0.21 moles HNO3

Molarity = moles / liter

1.5 M  = 0.21 moles / x

x = 0.21 moles / 1.5 M

x = 0.14 Liters HNO3

laik02076

Answer:

[tex]140mL[/tex]

Explanation:

[tex]M_1V_1=M_2V_2[/tex]

[tex]M_1=1.5M[/tex] molarity of nitric acid

[tex]V_1=?[/tex]  volume of nitric acid

[tex]M_2=7.0M[/tex] molarity of sodium hydroxide

[tex]V_2=30.0mL[/tex] volume of sodium hydroxide

[tex](1.5M)(V_1)=(7.0M)(30.0mL)[/tex]

[tex]V_1 =\frac{(7.0M)(30.0mL)}{1.5M}=140mL[/tex]

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