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andrezito [222]
1 year ago
11

A closed container contains 0.40 moles of argon gas at 25 °C and a pressure of 740 torr. The container is heated to 125 °C and t

he pressure increases. Estimate the number of moles of argon that must be released in order to drop the pressure back to near 740 torr while maintaining the temperature at 125 °C.
Chemistry
1 answer:
Ksivusya [100]1 year ago
7 0

The number of moles of argon that must be released in order to drop.

Solution:

Initial Temperature = 25°c = 298 K

Final Temperature =125 °c = 398 K

Initial Moles (n1) = 0.40 mole

Now,  Using the ideal gas law,

n1T1 = n2T2

0.400×298 = n2 × 398

n2 = 0.299 mol

Moles of Argon released

= 0.400-0.299

= 0.100 mol.

Pressure and force are related. That is using the physical equations if you know the other, you can calculate one using pressure = force/area. This pressure can be reported in pounds per square inch, psi, or Newtons per square meter N/m2. Kinetic energy causes air molecules to move faster. They hit the walls of the container more often and with greater force. The increased pressure inside the can may exceed the strength of the can and cause an explosion.

Learn more about The temperature here:-brainly.com/question/24746268

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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
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a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

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when PH = -㏒[H+]

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b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

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first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

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and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

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by substitution in H-H equation:

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we will use the H-H equation again as we have a buffer solution:

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first, we have to get moles HCIO = molarity * volume 

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                                                        = 0.0105 moles

then moles KOH = molarity * volume
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∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

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∴ the molarity of HCIO = moles HCIO remaining / total volume 

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and the molarity of KCIO = moles KCIO / total volume

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PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

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D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

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by using Kb expression:

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