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faltersainse [42]
3 years ago
8

If there are the same number of argon atoms in one sample and the same number of carbon atoms in another, then the value of the

ratio, mass of the carbon sample/the mass of argon sample would be?
a. 1.0
b. 3.3
c. 0.30
d. 12
e. 40
Chemistry
1 answer:
Art [367]3 years ago
8 0

Answer:

c. 0.30

Explanation:

One C atom has a mass of 12.0 u and one Ar atom has a mass of 40.0 u.

Let’s assume that there are <em>n</em> atoms of C and <em>n</em> atoms of Ar. Then

Mass of <em>n</em> C atoms = <em>n</em> × 12.0

Mass of <em>n</em> C atoms = 12.0<em>n</em> u

Mass of <em>n</em> Ar atoms = <em>n</em> × 40.0

Mass of <em>n</em> Ar atoms = 40.0<em>n</em> u

ratio = mass of C/mass of Ar

ratio = 12.0<em>n</em>/40.0<em>n </em>

ratio = 0.300

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Explanation:

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   \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}

                              = -log (1.7973 \times 10^{-4})

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Therefore, value of pK_{a} is 3.7454.

                             

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