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tankabanditka [31]
3 years ago
5

Whats the answer giving brainliest :)

Chemistry
1 answer:
MissTica3 years ago
5 0
It’s is c, if you oook it’s the only clear answer
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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
NeX [460]

Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

[SO^{2-}_{3}] = [Na_{2}SO_{3}] = 0.58 M

The sulphite anion will act as a base and react with H_{2}O to form HSO^{-}_{3} and OH^{-}.

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

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 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

So,

        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)

                x = 0.0003 M

So,   x = [HSO^{-}_{3}] = [OH^{-}] = 0.0003 M

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

                     = 0.579 M

Now, we will use [HSO^{-}_{3}] = 0.0003 M

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}}

                      = 7.14 \times 10^{-13}

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}

                     x = 0.0146 \times 10^{-6}

x = [OH^{-}] = [H_{2}SO_{3}] = 1.46 \times 10^{-8}

    [H^{+}] = \frac{10^{-14}}{[OH^{-}]}

                = \frac{10^{-14}}{0.0003}

                = 3.33 \times 10^{-11} M

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

5 0
3 years ago
what is the boiling point of a 0.321 m aqueous solution of NaCl. Enter your rounded answer with 3 decimal places.
Ronch [10]

Answer:

I believe the rounded place is 100.333.

3 0
3 years ago
Read 2 more answers
A 453 g piece of glass at 25.7∘C is left outside on a sunny day. How much heat must the glass absorb from the sun in order to re
Tju [1.3M]

Answer:

Q = 5555.6J

Explanation:

Mass of glass piece, m = 453g

initial temperature = 25.7°C

temperature to be attained = 40.3°C

⇒change in temperature, Δt = 40.3 - 25.7 = 14.6°C

specific heat of glass, s = 0.840J/g°C

Heat absorbed, Q = msΔt

⇒Q = 453×0.840×14.6 = 5555.592J

⇒<u>Q = 5555.6J</u> (rounded to the nearest tenth)

7 0
4 years ago
If you mix cu2+ with? a. nacl, b. nano2, c. h2o , arrange the solutions based on their absorption from highest frequency to lowe
erica [24]

The arrangement of the solutions based on their absorption from highest frequency to lowest frequency :

b.NaNO_{2} > c.H_{2} O > a.NaCl

<h3>What is absorption frequency?</h3>
  • The frequency of the molecular vibration that led to the absorption is the same as the absorption frequency of a basic IR absorption band.
  • In a way, an emission spectrum is the opposite of an absorption spectrum.
  • The discrepancies in the energy levels of each chemical element's orbitals correspond to absorption lines for each chemical element at various particular wavelengths.
  • Therefore, it is possible to identify the constituents in a gas or liquid using its absorption spectrum.
  • Absorption spectroscopy is most frequently used to measure infrared, atomic, visible, ultraviolet (UV), and x-ray waves.

Learn more about Absorption frequency here:

brainly.com/question/5032775

#SPJ4

3 0
1 year ago
“What types of friction occur between your bike tires and the ground when you ride over cement, through a puddle, and when you a
Kruka [31]
The right answer is= <span>rolling, fluid, and sliding</span>
4 0
3 years ago
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