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Ann [662]
2 years ago
7

How many molecules of co2 are in a 500. 0 ml container at 780 mm hg and 135°c? 8. 76 × 1021 molecules 9. 23 × 1021 molecules 5.

50 × 1021 molecules 2. 65 × 1022 molecules 2. 79 × 1022 molecules.
Chemistry
1 answer:
Aloiza [94]2 years ago
5 0

<u>Step 1:</u>

ok we have to use the formula PV=nRT

p=Pressure (must be converted to atm)= 780 mmHg

1 amt= 760 mmHg use this as a conversion factor

780 mmHg (1 atm/760 mmHg)= 1.026

V= Volume= 5.00 mL = o.5 L

n=number of moles which we have to find first

R= 0.0821

T(convert to Kelvins by adding 273.15 to the celsius temperature)= 135 C + 273.15= 408.15 k

Now plug in->

(1.026 atm)(o.5 L)= n(0.0821)(408.15 K)

(1.026 atm)(0.5 L)= n(33.509115)

(0.513)= n(33.509115)

n(number of moles)= 0.01532 mol

Now we have to convert to moles using Avagodro's number which states that 1 mol = 6.022 x 10^23 molecules or atoms

So 0.01532 mol (6.022 x 10^23 number of molesules)/ (1 mol) = 9.225704 x 10^21 = 9.226 x 10^21 colecules

Step 2

You must transfer pressure into pascals, 780 mm Hg = 103991 Pa

135*C = 408.15 k

then from the equation pV = nRt

n = pV / RT (T in Kelvins, V in M^3)

n = 103991 x 500 x 10^-6 / (8.314 x 408.15)= 0.015322 moles of N2

1 mol of everything is 6.022 x 10^23 particles, so 0.15322 moles is 0.15322 x 6.022 x 10^23 = 9.2269084 x 10^21 molecules

Explanation:

Hope this helps :)

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
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Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

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                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

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<h3>What is molarity?</h3>

Molarity of any solution is define as the number of moles of solute present in per liter of solution as;

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Moles of solute will be calculated as:

n = W/M, where

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On putting all values in the above equation, we get

V = (0.88) / (36.4) = 0.18 L

Hence required volume of water is 0.18L.

To know more about volume & concentration, visit the below link:

brainly.com/question/26762947

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