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8090 [49]
3 years ago
13

Correct first answer= Brianliest!PLEASE WOULD REALLY APPRECIATE IF YOU’D HELP ME OUT:’(

Chemistry
1 answer:
sveticcg [70]3 years ago
3 0

Answer:

The answer to your question is: 2 grams of methane

Explanation:

Data

V = 9.15 l

P = 1.77 atm

T = 57° C = 330 °K

Ar mass = 19 g

CH4 mass = ?

Formula

PV = nRT

Process

                     n = \frac{PV}{RT}

                      n = \frac{(1.77)(9.15)}{(0.082)(330)}

                      n = \frac{16.26}{27.06}

                             n = 0.6

Argon

                     40 g of Ar -------------------- 1 mol

                      19 g         ---------------------   x

                                 x = 0.475 mol of Ar

moles of CH4 = 0.6 - 0.475

                       = 0.125

                     

Methane

                            16 g of CH4 ---------------  1mol

                             x                 ----------------  0.125 mol

                            x = 2 grams

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Explanation:

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As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

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                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

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        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

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[SO^{2-}_{3}] = 0.58 - 0.0003

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The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              K_{b} = \frac{x^{2}}{0.0003 - x}

        K_{b} = \frac{K_{w}}{K_{a_{1}}}

                      = \frac{10^{-14}}{1.4 \times 10^{-2}}

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Therefore,  7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

                              = 0.00214 \times 10^{-13}

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                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

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