What volume of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid?
Nitric acid moles?
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To find the volume of lithium hydroxide needed to neutralize the nitric acid, we can use the equation:
[ \text{Volume of } \ce{LiOH} = \frac{{\text{Volume of } \ce{HNO3} \times \text{Normality of } \ce{HNO3}}}{{\text{Normality of } \ce{LiOH}}} ]
Plugging in the values:
[ \text{Volume of } \ce{LiOH} = \frac{{27.68 , \text{mL} \times 0.559 , \text{N}}}{{0.365 , \text{N}}} ]
[ \text{Volume of } \ce{LiOH} = \frac{{15.46712}}{{0.365}} ]
[ \text{Volume of } \ce{LiOH} \approx 42.41 , \text{mL} ]
So, approximately 42.41 mL of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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