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Tasya [4]
3 years ago
7

A syringe contains 610 mL of CO at 310 K and 1.5 atm pressure. A second syringe contains 520 mL of N2 at 325 K and 3.5 atm. What

is the final pressure if the contents of these two syringes are injected into 2.00 L container at 10.0C
Chemistry
1 answer:
scoray [572]3 years ago
5 0

Answer:

P = 1.21atm

Explanation:

Using PV = nRT, moles of both syringes is:

Moles CO:

n = PV / RT

n = 1.5atm*0.610L / 0.082atmL/molK*310K

n = 0.0360 moles

Moles N₂:

n = PV / RT

n = 3.5atm*0.520L / 0.082atmL/molK*325K

n = 0.0683 moles.

As in the container you mix both gases, moles in the container are:

n = 0.0360 + 0.0683 = <em>0.1043 moles</em>

Conditions of the container are:

V = 2.00L; T = 273.15K + 10°C = 283.15K; n = 0.1043 moles.

Thus, pressure is:

P = nRT / V

P = 0.1043mol*0.082atmL/molK*283.15K / 2.00L

<h3>P = 1.21atm</h3>
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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
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Explanation:

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    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

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          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

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