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denis23 [38]
3 years ago
11

Nitric acid is usually purchased in concentrated form with a 70.3% HNO3HNO3 concentration by mass and a density of 1.41 g/mLg/mL

. Part A How much of the concentrated stock solution in milliliters should you use to make 2.0 LL of 0.500 MM HNO3HNO3 ?
Chemistry
1 answer:
vodomira [7]3 years ago
6 0

Answer:

V = 63.57 mL

Explanation:

In this case, to get the mL of the stock nitric acid we need to know first the concentration in mol/L so it can match the units of the desired concentration, in this case, 0.5 mol/L

To get the concentration of the stock solution, we should use the following expression:

M = % * d * 1000 / MM * 100

The molar mass of the nitric acid (HNO₃) is:

MM = 1 + 14 + 3(16) = 63 g/mol

the concentration of the acid is:

M = 70.3 * 1.41 * 1000 / 63 * 100

M = 15.73 M

Now that we know the concentration of the solution, we use the following expression to get the volume:

M1*V1 = M2*V2

Replacing the values and solving for V1 we have:

V1 = M2*V2 / M1

V1 = 2 * 0.5 / 15.73

V1 = 0.06357 L

In mL it will be:

V1 = 0.06357 * 1000 = 63.57 mL

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A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

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3 years ago
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