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3 years ago

What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?

Chemistry

1 answer:

IRINA_888 [86]3 years ago

5 0

<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

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

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

= 17.5 g ÷ 39.997 g/mol

= 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

= 0.4375 moles ÷ 0.250 M

= 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

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Answer:

$\boxed {\boxed {\sf 0.80 \ mol\ F}}$

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

$\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

$4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

Flip the ratio so the units of atoms of fluorine cancel each other out.

$4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}$

$4.8 \times 10^{23} *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }$

Condense into 1 fraction.

$\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F$

Divide.

$0.7970773829 \ mol \ F$

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

$0.80 \ mol \ F$

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

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