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Luda [366]
3 years ago
10

In a titration, a few drops of an indicator are added to a flask containing 35.0 milliiters of HNO3(aq) of unknown concentration

. After 30.0 millliters of 0.15 M NaOH(aq) solution is slowly added o the flask, the indicator changes color, showing the acid is neutralized. Which choice is a correct numerical setup for calculating the concentration of the HNO3(aq) solution
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

0.13M HNO3 is the concentration of the unknown

Explanation:

Nitric acid, HNO3, reacts with NaOH as follows:

HNO3 + NaOH  → H2O + NaNO3

<em>Where 1mol HNO3 reacts with 1mol NaOH</em>

<em />

To find the concentration of HNO3 we must find, as first, the moles NaOH added = Moles HNO3 in the solution:

30mL * (1L / 1000mL) * (0.15mol / 1L) = 0.0045 moles HNO3

In 35.0mL = 0.035L:

0.0045 moles HNO3 / 0.035L =

0.13M HNO3 is the concentration of the unknown

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Answer:

<h2>0.92 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

mass = 16.1 g

volume = 17.5 mL

density =  \frac{16.1}{17.5}  = 0.92 \\

We have the final answer as

<h3>0.92 g/mL</h3>

Hope this helps you

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