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Leokris [45]
3 years ago
13

What is the [SO32-) in the 0.080 M H2SO3 .080 M Solution?

Chemistry
1 answer:
slega [8]3 years ago
4 0

Answer:

0.080 M

Explanation:

From the question given above, the following data were obtained:

Concentration of H₂SO₃ = 0.080 M

Concentration of SO₃²¯ =…?

The concentration of SO₃²¯, can be obtained as follow:

H₂SO₃ (aq) <==> 2H⁺ (aq) + SO₃²¯ (aq)

From the balanced equation for above,

1 mole of H₂SO₃ produced 1 mole of SO₃²¯.

Therefore, 0.080 M H₂SO₃ will also produce 0.080 M SO₃²¯.

Thus, the concentration of SO₃²¯ is 0.080 M

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When Niels Bohr refined the model of an atom,what new idea did he include?
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3 years ago
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3 years ago
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

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Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

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Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

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Answer: The answer is D :)

Explanation:

3 0
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